Why can ALL quadratic equations be solved by the quadratic formula?












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In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.



Why can I use $x = dfrac{-bpm sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?










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    Look at Wikipedia.
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    – t.b.
    Jul 3 '11 at 16:16






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    This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
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    – KCd
    Jul 3 '11 at 17:31






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    @KCd: Not to mention the existence of a square root.
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    – Phira
    Jul 3 '11 at 18:56






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    KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
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    – Mark Bennet
    Jul 3 '11 at 19:16






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    @idonno I admire your curiosity! If only precalc students were more inquisitive!
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    – Pedro Tamaroff
    Feb 23 '12 at 3:38
















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In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.



Why can I use $x = dfrac{-bpm sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?










share|cite|improve this question











$endgroup$








  • 13




    $begingroup$
    Look at Wikipedia.
    $endgroup$
    – t.b.
    Jul 3 '11 at 16:16






  • 95




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    This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
    $endgroup$
    – KCd
    Jul 3 '11 at 17:31






  • 28




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    @KCd: Not to mention the existence of a square root.
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    – Phira
    Jul 3 '11 at 18:56






  • 18




    $begingroup$
    KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
    $endgroup$
    – Mark Bennet
    Jul 3 '11 at 19:16






  • 26




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    @idonno I admire your curiosity! If only precalc students were more inquisitive!
    $endgroup$
    – Pedro Tamaroff
    Feb 23 '12 at 3:38














256












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In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.



Why can I use $x = dfrac{-bpm sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?










share|cite|improve this question











$endgroup$




In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.



Why can I use $x = dfrac{-bpm sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?







algebra-precalculus polynomials quadratics






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edited Mar 8 '16 at 16:21









Andreas Caranti

56.5k34395




56.5k34395










asked Jul 3 '11 at 16:10









idonnoidonno

1,72282020




1,72282020








  • 13




    $begingroup$
    Look at Wikipedia.
    $endgroup$
    – t.b.
    Jul 3 '11 at 16:16






  • 95




    $begingroup$
    This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
    $endgroup$
    – KCd
    Jul 3 '11 at 17:31






  • 28




    $begingroup$
    @KCd: Not to mention the existence of a square root.
    $endgroup$
    – Phira
    Jul 3 '11 at 18:56






  • 18




    $begingroup$
    KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
    $endgroup$
    – Mark Bennet
    Jul 3 '11 at 19:16






  • 26




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    @idonno I admire your curiosity! If only precalc students were more inquisitive!
    $endgroup$
    – Pedro Tamaroff
    Feb 23 '12 at 3:38














  • 13




    $begingroup$
    Look at Wikipedia.
    $endgroup$
    – t.b.
    Jul 3 '11 at 16:16






  • 95




    $begingroup$
    This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
    $endgroup$
    – KCd
    Jul 3 '11 at 17:31






  • 28




    $begingroup$
    @KCd: Not to mention the existence of a square root.
    $endgroup$
    – Phira
    Jul 3 '11 at 18:56






  • 18




    $begingroup$
    KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
    $endgroup$
    – Mark Bennet
    Jul 3 '11 at 19:16






  • 26




    $begingroup$
    @idonno I admire your curiosity! If only precalc students were more inquisitive!
    $endgroup$
    – Pedro Tamaroff
    Feb 23 '12 at 3:38








13




13




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Look at Wikipedia.
$endgroup$
– t.b.
Jul 3 '11 at 16:16




$begingroup$
Look at Wikipedia.
$endgroup$
– t.b.
Jul 3 '11 at 16:16




95




95




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This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
$endgroup$
– KCd
Jul 3 '11 at 17:31




$begingroup$
This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.
$endgroup$
– KCd
Jul 3 '11 at 17:31




28




28




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@KCd: Not to mention the existence of a square root.
$endgroup$
– Phira
Jul 3 '11 at 18:56




$begingroup$
@KCd: Not to mention the existence of a square root.
$endgroup$
– Phira
Jul 3 '11 at 18:56




18




18




$begingroup$
KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
$endgroup$
– Mark Bennet
Jul 3 '11 at 19:16




$begingroup$
KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.
$endgroup$
– Mark Bennet
Jul 3 '11 at 19:16




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@idonno I admire your curiosity! If only precalc students were more inquisitive!
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– Pedro Tamaroff
Feb 23 '12 at 3:38




$begingroup$
@idonno I admire your curiosity! If only precalc students were more inquisitive!
$endgroup$
– Pedro Tamaroff
Feb 23 '12 at 3:38










22 Answers
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I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.



Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)



We want to solve the equation
$$ax^2+bx+c=0,$$
where $a ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.



Instead, multiply both sides by $4a$. We obtain the equivalent equation
$$4a^2x^2 +4abx+4ac=0.tag{1}$$
Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
$$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
So our equation can be rewritten as
$$(2ax+b)^2 -b^2+4ac=0 tag{2}$$
or equivalently
$$(2ax+b)^2=b^2-4ac. tag{3}$$
Now it's all over. We find that
$$2ax+b=pmsqrt{b^2-4ac} tag{4}$$
and therefore
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}. tag{5}$$

No fractions until the very end!



Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.



Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.






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    I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
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    – Pete L. Clark
    Jul 3 '11 at 20:31






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    @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
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    – André Nicolas
    Jul 3 '11 at 20:53








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    I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
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    – Pete L. Clark
    Jul 3 '11 at 21:28








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    @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
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    – André Nicolas
    Jul 3 '11 at 21:37






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    In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
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    – André Nicolas
    May 14 '14 at 15:50



















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Here is a slightly less ad-hoc approach to deriving the formula.



You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,times,div$ and $sqrt[n]{}$.



But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.



Intuitively because you know that $(r_1+r_2)=-frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=frac{-b}{2a}pmfrac{E}2$$



All we need to do then, is express $E=(r_1-r_2)$ using $+,-,times,div,sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,times,div$ and $a,b,c$ could possible be.



Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.



In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.



We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$



Evidently, now we have that $sqrt{D}=a(r_1-r_2)=aE$ and so $E=frac{sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=frac{-b}{2a}pmfrac{E}{2}$$ to $$r_1,r_2=frac{-b}{2a}pmfrac{sqrt{D}}{2a}$$ and finally $$r_1,r_2=frac{-bpmsqrt{b^2-4ac}}{2a}$$





The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.



What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of ${dfrac {p(a,b,c)}{q(a,b,c)}}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).



For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,times,div,sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.






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    This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
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    – Pedro Tamaroff
    May 2 '12 at 23:03








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    @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
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    – Vladimir Sotirov
    Apr 19 '14 at 21:02






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    @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
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    – Anastasiya-Romanova 秀
    Apr 20 '14 at 8:59



















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Complete the square,



If you dare,



And the Answer



Will be there!






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    What value does this add?
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    – Peter Taylor
    Jul 4 '11 at 12:29






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    @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
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    – Mike Jones
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Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.



The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{bover a}x+{cover a}=0$. Now, that $bover a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{bover 2a})^2 = x^2+{bover a}x + {b^2over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $cover a$. A correction of $({cover a}-{b^2over 4a^2})$ will do this; we get $(x+{bover 2a})^2+({cover a}-{b^2over 4a^2}) = 0$.



But this is exactly what we wanted; we can move that second term over to the right and get $(x+{bover 2a})^2 = {b^2over 4a^2}-{cover a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4acover 4a^2}$ - just multiply the numerator and denominator of $cover a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{bover 2a} = pm sqrt{b^2-4acover 4a^2} = pm {sqrt{b^2-4ac}oversqrt{4a^2}} = {pmsqrt{b^2-4ac}over 2a}$. The last step is to subtract $bover 2a$ from both sides, finally giving the familiar: $x = {-bpmsqrt{b^2-4ac}over 2a}$






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    Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
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    – André Nicolas
    Jul 3 '11 at 17:01










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    If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
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    – dbasnett
    Jul 3 '11 at 17:51






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    @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
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    – André Nicolas
    Jul 3 '11 at 18:03






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    And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
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    – dbasnett
    Jul 3 '11 at 18:07






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    It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
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    – Pete L. Clark
    Jul 3 '11 at 22:04





















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Proof without words.



completing the square



This one shows that $$ax^2+bx+c=aleft(x+dfrac b{2a}right)^2+c-dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.



Credits to LucasVB.



I hope this helps.

Best wishes, $mathcal H$akim.






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    The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity
    $$
    aX^2+bX+c=aleft(X-frac{-b+sqrt{b^2-4ac}}{2a}right)left(X-frac{-b-sqrt{b^2-4ac}}{2a}right)
    $$
    holds for every $a$, $b$ and $c$.






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      In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
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      – Martin Brandenburg
      Jul 26 '14 at 9:05












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      Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
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      – Carsten S
      Dec 10 '16 at 12:00



















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    Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.



    Try working backwards!



    Assuming that the quadratic formula holds true,$$x = {-b pmsqrt{b^2 - 4ac} over 2a} $$Isolate $x$.$$begin{align}2ax &=& -b pm sqrt{b^2 - 4ac} \ 2ax + b & = &pmsqrt{b^2 - 4ac} \ (2ax + b)^2 & = & b^2 - 4ac \ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \ 4a^2x^2 + 4abx + 4ac & = & 0 \ ax^2 + bx + c & = & 0 end{align}$$
    Note that we divided both sides by $4a$ in the last step assuming a $ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}cdots a_0x^0$, $a_n ne 0$.



    Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!






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      From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
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      – Deepak Gupta
      Sep 18 '15 at 7:20



















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    First, let's examine an analogous simpler special case. Why does the difference of squares formula $rm: x^2 - a^2 = (x-a) (x+a):$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $rm (x-a) (x+a) = x^2 - a x + x a - a^2 $ which indeed equals $rm x^2 - a^2 $ as long as $rm a x = x a $ for all $rm:x:,:$ i.e. as long as $rm:a:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.



    In fact, if we consider both $rm:x:$ and $rm:a:$ as indeterminates, then we can specialize the "generic" factorization $rm: x^2 - a^2 = (x-a) (x + a):$ in the ring $rm:mathbb Z[x,a]:$ to any commutative ring by using an evaluation homomorphism mapping $rm:x,a:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.



    A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $rm:2:a:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $rm:e = 1/(2a), d^2 = b^2 - 4ac:,:$ so the proofs works naturally in the ring $rm:mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac):.:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.



    Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $rm det (1+AB)=det (1+BA) $ that proceeds by universally cancelling $rm det A $ from the $rm det $ of $rm (1+A B) A = A (1+B A),$ in $,rmBbb Z[a_{,ij},b_{,ij}],$ where the matrix entries are indeterminates $rm,a_{,ij},b_{,ij}.,$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).



    REMARK $ $ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $rm:x^2 = 1:$ has roots $rm:pm1,:pm3 in mathbb Z/8:,:$ the ring of integers modulo $8:.:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $rm xy = 0 Rightarrow x=0 or y=0:.;$ Indeed, a polynomial $rm f(x)in D[x] $ has at most $rm deg f $ roots in the ring $rm:D $ iff $rm D:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $rm mathbb Z/m $ by showing that, given any $rm:f(x):$ with more roots than its degree, we can quickly compute a nontrivial factor of $rm:m:$ via a $rm:gcd:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $rm:m:$ by searching for a nontrivial square root in $rm: mathbb Z/m:,:$ e.g. a square root of $1$ that is not $:pm 1$.






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      Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
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      – Pete L. Clark
      Jul 3 '11 at 23:30






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      @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
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      – Bill Dubuque
      Jul 3 '11 at 23:37






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      @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
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      – Bill Dubuque
      Jul 3 '11 at 23:42








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      $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
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      – Pete L. Clark
      Jul 3 '11 at 23:56






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      @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
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      – Bill Dubuque
      Jul 4 '11 at 14:05



















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    The formula comes from the technique known as completing the square.






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      I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.




      Consider any quadratic equation



      $$ax^{2}+bx+c=0,qquad (aneq 0).qquad (1)$$



      Multiplying both sides by $1/a$ we get
      the equivalent equation



      $$x^{2}+frac{b}{a}x+frac{c}{a}=0.$$



      We are going to show that it is possible to find $h$ and $alpha $ such that



      $$x^{2}+frac{b}{a}x+frac{c}{a}=(x+h)^{2}-alpha .$$



      Expanding the RHS gives



      $$x^{2}+frac{b}{a}x+frac{c}{a}=x^{2}+2hx+h^{2}-alpha .$$



      This means, applying the method of undetermined coefficients, that



      $$left{ begin{array}{l} frac{b}{a}=2h \ frac{c}{a}=h^{2}-alpha. end{array} right. $$



      Hence



      $$left{ begin{array}{l}h=frac{b}{2a} \ alpha=h^{2}-frac{c}{a}=frac{b^{2}}{4a^{2}}-frac{c}{a}=frac{b^{2}-4ac}{4a^{2}}.end{array}right. qquad (2)$$



      In this way the given equation is reduced to the binomial equation in
      $x+h$



      $$left( x+hright) ^{2}-alpha =0qquad text{equivalent to}qquad left( x+hright) ^{2}=alpha $$



      and thus it is satisfied when



      $$x+h=sqrt{alpha }qquad text{or}qquad x+h=-sqrt{alpha },$$



      i.e. when



      $$x=-h+sqrt{alpha }qquad text{or}qquad x=-h-sqrt{alpha }.$$



      Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $alpha $ by their expressions given by $(2)$, yields



      $$x_{1}=frac{-b+sqrt{b^{2}-4ac}}{2a}qquad text{or}qquad x_{2}=frac{-b-sqrt{b^{2}-4ac}}{2a}.qquad (3)$$




      As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.






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        Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).

        So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)






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          I thought that was implied by "degree two"!
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          – The Chaz 2.0
          Jul 4 '11 at 5:40



















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        This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $mathbb{R}$ or $mathbb{Q}$).



        Step 1. The standard identity
        begin{equation}
        (x + y)^2 = (x-y)^2 + 4xy qquad qquad (1)
        end{equation}



        Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = frac{1}{2}(x+y) + (x-y)$ and $y = frac{1}{2}(x+y) + (x-y)$.



        Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a not= 0$. Dividing by $a$ yields $x^2 + frac{b}{a}x +frac{c}{a} = 0$ or $x(x + frac{b}{a}) = -frac{c}{a}$. Setting $y = x + frac{b}{a}$, we get $xy= -frac{c}{a}$ and $x - y = -frac{b}{a}$.
        The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.



        I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.






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          You could also prove it using Tschirnhaus Transformation.



          Let $,x=y-frac{b}{2a}$.$,$ Then $,ax^2+bx+c=0,$ becomes
          $$begin{align}ay^2-frac{b^2}{4a}+c&=0\
          iff y^2&=frac{b^2-4ac}{4a^2}\
          iff y&=frac{pmsqrt{b^2-4ac}}{2a}\
          iff x&=frac{pmsqrt{b^2-4ac}}{2a}-frac{b}{2a}\
          iff x&=frac{-bpmsqrt{b^2-4ac}}{2a}end{align}$$





          In general, the substitution for $, a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0,$ is $,x=y-frac{a_{n-1}}{na_n}$,



          which leaves the polynomial without a term of degree $n-1$.






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            Hardly adequate. Clearly the OP was looking for more than a standard derivation.
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            – Daniel W. Farlow
            Feb 8 '15 at 4:19






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            @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
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            – user26486
            Feb 8 '15 at 4:48








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            I agree with @induktio
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            – fancynancy
            Feb 8 '15 at 5:28






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            I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
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            – user26486
            Feb 19 '15 at 15:01



















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          Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.



          Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $alpha$ and $beta$.
          Now, $$~~~~~~~~~~~~~~~~~~~~~x-alpha = x-beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$
          $$~~~~~~~~~~~~~~~~~~~~~k(x-alpha)(x-beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
          Equating equation 1 and 2 (k is a constant),
          $$ax^{2}+b{x}+c=k(x-alpha)(x-beta)$$
          $$ax^{2}+b{x}+c=k(x^{2}-alpha x-beta x+alpha beta)$$
          $$ax^{2}+b{x}+c=kx^{2}-k(alpha+beta )x+kalpha beta)$$
          Comparing both sides, we get $$a=k~;~b=-k(alpha +beta)~;~c=kalpha beta$$
          From this, we get $$alpha + beta = frac{-b}{a}~~;~~~alpha beta = frac{c}{a}$$



          Now, to get the value of $alpha$, we follow the following procedure :
          First we take out the value of $alpha - beta$, so that we can eliminate one term and find out the value of another.
          $$(alpha-beta)^{2} = alpha ^{2}+ beta ^{2} - 2 alpha beta$$
          Now we'll add $4 alpha beta $ on both the sides
          $$(alpha-beta)^{2} +4 alpha beta = alpha ^{2}+ beta ^{2} + 2 alpha beta$$
          $$(alpha-beta)^{2} +4 alpha beta = (alpha + beta )^{2} $$
          $$(alpha-beta)^{2} = (alpha + beta )^{2} -4 alpha beta $$
          $$alpha-beta = pm sqrt{(alpha + beta )^{2} -4 alpha beta } $$
          Substituting the values of $alpha + beta$ and $alpha beta$, we get,
          $$alpha-beta = pm sqrt{(frac{-b}{a} )^{2} -frac{4c}{a} } $$
          $$alpha-beta = pm sqrt{frac{b^{2}-4ac}{a^{2}} } $$ or
          $$~~~~~~~~~~~~~~~~~~~~~alpha-beta = frac{pm sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
          Adding $Eq^{n} (2)~and~(3)$, we get,
          $$2 alpha = frac{-b pm sqrt{b^{2}-4ac}}{a}$$
          $$alpha = frac{-b pm sqrt{b^{2}-4ac}}{2a}$$






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          • 2




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            $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
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            – miracle173
            Nov 25 '14 at 5:16












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            you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
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            – user173094
            Nov 25 '14 at 11:35



















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          I'll just let the algebra speak for itself.



          For $ a ne 0 $



          If



          $ax^2 = bx + c $



          Then



          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



          PROOF:



          $ 4aax^2 = 4abx + 4ac $



          $(2ax)^2 + b^2 = 4abx + 4ac + b^2 $



          $(2ax)^2 - 4abx + b^2 = b^2 + 4ac $



          $ (2ax - b)^2 = b^2 + 4ac $



          $ 2ax - b = pm sqrt{b^2 + 4ac} $



          $2ax = b pm sqrt{b^2 + 4ac} $



          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



          Q.E.D.



          SHORTCUT



          Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.



          For $ a ne 0 $



          If



          $ ax^2 + bx + c = 0 $



          Then



          $ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} $



          Make the replacements in the 'if'



          $ ax^2 + (-b)x + (-c) = 0 $



          $ ax^2 = bx + c $



          Make the replacements in the 'then'



          $x = frac{-(-b) pm sqrt{(-b)^2 - 4a(-c)}}{2a} $



          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



          Q.E.D.






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            The proof of the quadratic formula takes advantage of completing the square.
            $$ax^2 + bx + c = 0, a neq 0, a, b, c in mathbb R$$
            $$ax^2 + bx = -c$$
            $$x^2 + dfrac{b}{a}x = -dfrac{c}{a}$$
            $$x^2 + dfrac{b}{a}x + left(dfrac{b}{2a}right)^2 = -dfrac{c}{a} + left(dfrac{b}{2a}right)^2$$
            $$left(x+dfrac{b}{2a}right)^2 = -dfrac{c}{a} + dfrac{b^2}{4a^2}$$
            $$left(x+dfrac{b}{2a}right)^2 = -dfrac{4ac}{4a^2} + dfrac{b^2}{4a^2}$$
            $$left(x+dfrac{b}{2a}right)^2 = dfrac{-4ac+b^2}{4a^2}$$
            $$left(x+dfrac{b}{2a}right)^2 = dfrac{b^2-4ac}{4a^2}$$
            $$x+dfrac{b}{2a} = pm sqrt{dfrac{b^2-4ac}{4a^2}}$$
            $$x+dfrac{b}{2a} = pm dfrac{sqrt{b^2-4ac}}{2a}$$
            $$x=-dfrac{b}{2a} pm dfrac{sqrt{b^2-4ac}}{2a}$$
            $$boxed{x=dfrac{-b pm sqrt{b^2-4ac}}{2a}}$$
            This concludes the proof.






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              The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=pm sqrt{a}$.



              The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+frac{b}{a} x + frac{c}{a}=0$. Writing $p=frac{b}{a}$, $q=frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-frac{p}{2}$, then a little calculation shows $y^2 - frac{p^2}{4}+q=0$.



              $$Rightarrow ~ y=pm sqrt{frac{p^2}{4}-q} ~ Rightarrow ~ x = -frac{p}{2} pm sqrt{frac{p^2}{4}-q} = frac{-b pm sqrt{b^2-4ac}}{2a}.$$



              Geometrically, $x=y-frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.



              enter image description here



              This method also works for polynomials of higher degrees: Tschirnhaus transformation.






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                it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
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                – user153330
                Mar 21 '16 at 20:04





















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              I've seen several "proofs" of this formula, all of which proceed by a sequence of randomly-chosen algebraic operations and end at the famous formula. Of course, no normal human being would ever pluck these out of thin air like that.



              For me, the realization was this: Consider the formula



              $$(x - alpha)(x - beta) = 0$$



              If $x=alpha$, then $x-alpha$ is obviously zero. Whatever $x-beta$ happens to be, it gets multiplied by zero, so the result is still zero. So in other words, one solution to this equation is $x=alpha$. By a similar line of reasoning, $x=beta$ is also a solution. So here we have purposely constructed an equation that has exactly two solutions: $alpha$ and $beta$ (which can be whatever we choose them to be).



              Now let's open the brackets:



              $$(x-alpha)(x-beta) = 0$$
              $$x(x-beta) - alpha(x-beta) = 0$$
              $$x^2 - beta x - alpha x + alpha beta = 0$$
              $$x^2 - (alpha + beta)x + (alpha beta) = 0$$



              Now suppose we have an equation that looks like



              $$x^2 + Ax + B = 0$$



              (Notice that this isn't the usual general quadratic equation. I've assumed that the $x^2$ has a coefficient of 1.)



              If we line the two equations up, it becomes clear that



              $$A = -(alpha + beta)$$
              $$B = (alpha beta)$$



              This is the direct relationship between the coefficients we can see, and the solutions we want to find. Now, how to figure out the solutions from the coefficients?



              We know that $-A$ is the sum of the solutions, and $B$ is the product of the solutions. But what are the solutions themselves? Hmm.



              Well, if $-A$ is the sum, then $-A/2$ is the arithmetic mean of the solutions. I.e., it's exactly half way between the two solutions. Great, if we can just figure out how far apart the solutions are, we could exactly compute them!



              So we have the sum and the product, but we want the difference. Hmm. OK. But how do we do that?



              We've got $alpha + beta$ and we've got $alpha beta$, but we want $alpha - beta$. Tricky…



              The solution is Black Magic. Observe: The thing we have is $alpha + beta$, and



              $$(alpha + beta)^2 = alpha^2 + 2 alpha beta + beta^2$$



              The thing we want is $alpha - beta$, and



              $$(alpha - beta)^2 = alpha^2 - 2 alpha beta + beta^2$$



              The difference between these two expressions is exactly $4alphabeta$which we can compute! If $B = alphabeta$, then $4B = 4alphabeta$. What I am saying is that



              $$(alpha - beta)^2 = (alpha + beta)^2 - 4alphabeta = A^2 - 4B$$



              It therefore follows that



              $$alpha - beta = sqrt{A^2 - 4B}$$



              Recalling that $-A$ is the sum of the solutions and so $-A/2$ is the average, we have



              $$alpha = -frac{A}{2} - frac{sqrt{A^2 - 4B}}{2}$$
              $$beta = -frac{A}{2} + frac{sqrt{A^2 - 4B}}{2}$$



              In other words, half the sum minus half the difference gives the lower solution, and half the sum plus half the difference gives the upper solution.



              $$x = frac{-A pm sqrt{A^2 - 4B}}{2}$$



              Notice that this doesn't look quite like the familiar formula, because we assumed that the $x^2$ coefficient equals exactly 1. But notice that if we divide all the coefficients by the leading one, we can get this form!



              $$ax^2 + bx + c = 0$$
              $$x^2 + frac{b}{a}x + frac{c}{a} = 0$$
              $$A = frac{b}{a}$$
              $$B = frac{c}{a}$$



              Substitute these into our previous formula, and (trust me) the familiar quadratic solution formula pops out.





              Notice that we can do this trick for a polynomial of any degree!



              $$(x - alpha)(x - beta)(x - gamma) = 0$$



              This gives a cubic equation with known solutions. Opening the brackets,



              $$x^3 - (alpha + beta + gamma)x^2 + (alphabeta + alphagamma + betagamma)x - (alpha beta gamma) = 0$$



              Given the equation



              $$x^3 + Ax^2 + Bx + C = 0$$



              we get



              $$A = -(alpha + beta + gamma)$$
              $$B = (alphabeta + alphagamma + betagamma)$$
              $$C = -(alphabetagamma)$$



              The trouble is, now we need to somehow compute $alpha$, $beta$ and $gamma$ given only $A$, $B$ and $C$. And it turns out this is way harder than last time. (Indeed, I have no damn idea how to do it! Computing the average doesn't really help this time.)






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                Graphical Approach
                The following method actually requires some knowledge about the graph of parabola and calculus. This is just another way to view the formula.
                For $a>0,,
                f^{prime}(x)=0,left[x=-frac{b}{2a}right]$ would give you the minima. Now substitute this in the expression to see whether it attains any negative value. If it is positive here then it wont have any real roots for obvious reasons given that you have sufficient knowledge about functions.
                Now if the function has real roots then the arithmetic mean of the roots would give the axis of parabola. The roots are symmetrical about this line so if we somehow find the difference between the roots and add half of it to $-frac{b}{2a}$ then we would get one root and subtracting half of difference would give the other root.We know the sum of roots is $-frac{b}{a}$ and product of roots is $frac{c}{a}$. So
                begin{equation}
                F+G=-frac{a}{b},,FG=frac{c}{a}$
                end{equation}
                therefore, $(F-G)^2 = left(-frac{a}{b}right)^2 -4frac{c}{a}$
                taking root,
                $|F-G|=left[left(-frac{a}{b}right)^2 -4frac{c}{a}right]^{frac{1}{2}}$
                therefore the roots are
                $-frac{b}{a} + frac{|F-G|}{2},-frac{b}{a} - frac{|F-G|}{2}$ enter image description here
                Substituting the values you would get the formula. I think this graphical way is a much interesting way of looking at the formula as algebra here can be too abstract for people new to it.






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                • $begingroup$
                  It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
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                  – David K
                  Mar 8 '16 at 16:03



















                2












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                Our approach exploits the fact that any differentiable function $f:mathbb{R} rightarrow mathbb{R}$ is uniquely specified by its derivative and a point on its graph. The quadratic polynomial $P(x) = ax^2 + bx + c$ where $a neq 0$ satisfies $P(0) = c$ and $P'(x) = 2ax + b$.



                We have $P'(x) = 2ax + b = a(2(x+frac {b}{2a}))$. We can deduce from this that $g(x) = acdot(x+frac{b}{2a})^2 + C$ has the same derivative as $P(x)$. To make it the same function, it must have the same value at a point on its graph, say at $x=0$.



                In other words, we must make $g(0) = frac{b^2}{4a} + C = c$ which can be done if we let $C = c - frac{b^2}{4a}$. We have thus shown $ax^2 + bx + c = acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a}$



                To derive the quadratic formula, $acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a} = 0 $. This can be solved using elementary algebraic manipulations, from which we get the desired $x = frac{-bpm sqrt{b^{2} - 4 ac}}{2a}$






                share|cite|improve this answer









                $endgroup$





















                  1












                  $begingroup$

                  If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant is called completing the square). The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.



                  So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B pm sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.






                  share|cite|improve this answer









                  $endgroup$





















                    0












                    $begingroup$

                    There are essentially two questions being asked here:
                    1. Why does the quadratic formula apply to ALL quadratic equations?
                    2. Why is the quadratic formula correct / How can I prove this?



                    Most of the answers have focussed on the second version of this question, and more specifically, at simply coming up with different proofs, rather than instructing how to come up with one yourself. Fair enough, since these proofs have been incredibly instructive to many readers including me.



                    But it does leave a bit of a gap. I'll address the first part of that gap by answering the first question: why does the quadratic formula actually apply to all quadratic equations?



                    That's because the quadratic formula expresses its solution without assuming any information about the actual value of any of the constants a, b, and c. Therefore, it can be applied to a quadratic equation with any values for these three constants.



                    This is what mathematics is all about: finding general solutions to things so that you can use the same solution method for a wide array of problems.



                    The same could be said about why the number "3" can be used to describe any group of that size; a group of three sheep, three people, three bricks, whatever. The whole reason why that concept was invented and why it survived the passage of time is because it's so useful.



                    It's useful because it can be applied to a wide array of different things, rather than requiring a re-invention of the wheel for every different kind of object. This can, of course, only be done by presuming as little as possible to whatever the concept can be applied to.



                    And that's why concepts like the quadratic formula show up in your maths classes in the first place. They solve a very important problem because many real-life things can be modeled as polynomials, especially the lower ones like the linear and quadratic, so being able to solve such a thing with just one "wheel" is incredibly important and useful.



                    Hopefully that adds a little perspective on the general applicability and consequent utility of the quadratic formula, the former essentially being what the question was inquiring under its most literal interpretation.






                    share|cite|improve this answer









                    $endgroup$












                      protected by Zev Chonoles Jan 19 '16 at 18:14



                      Thank you for your interest in this question.
                      Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



                      Would you like to answer one of these unanswered questions instead?














                      22 Answers
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                      22 Answers
                      22






                      active

                      oldest

                      votes









                      active

                      oldest

                      votes






                      active

                      oldest

                      votes









                      495












                      $begingroup$

                      I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.



                      Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)



                      We want to solve the equation
                      $$ax^2+bx+c=0,$$
                      where $a ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.



                      Instead, multiply both sides by $4a$. We obtain the equivalent equation
                      $$4a^2x^2 +4abx+4ac=0.tag{1}$$
                      Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
                      $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
                      So our equation can be rewritten as
                      $$(2ax+b)^2 -b^2+4ac=0 tag{2}$$
                      or equivalently
                      $$(2ax+b)^2=b^2-4ac. tag{3}$$
                      Now it's all over. We find that
                      $$2ax+b=pmsqrt{b^2-4ac} tag{4}$$
                      and therefore
                      $$x=frac{-bpmsqrt{b^2-4ac}}{2a}. tag{5}$$

                      No fractions until the very end!



                      Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.



                      Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.






                      share|cite|improve this answer











                      $endgroup$









                      • 57




                        $begingroup$
                        I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 20:31






                      • 76




                        $begingroup$
                        @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 20:53








                      • 9




                        $begingroup$
                        I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 21:28








                      • 38




                        $begingroup$
                        @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 21:37






                      • 5




                        $begingroup$
                        In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                        $endgroup$
                        – André Nicolas
                        May 14 '14 at 15:50
















                      495












                      $begingroup$

                      I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.



                      Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)



                      We want to solve the equation
                      $$ax^2+bx+c=0,$$
                      where $a ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.



                      Instead, multiply both sides by $4a$. We obtain the equivalent equation
                      $$4a^2x^2 +4abx+4ac=0.tag{1}$$
                      Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
                      $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
                      So our equation can be rewritten as
                      $$(2ax+b)^2 -b^2+4ac=0 tag{2}$$
                      or equivalently
                      $$(2ax+b)^2=b^2-4ac. tag{3}$$
                      Now it's all over. We find that
                      $$2ax+b=pmsqrt{b^2-4ac} tag{4}$$
                      and therefore
                      $$x=frac{-bpmsqrt{b^2-4ac}}{2a}. tag{5}$$

                      No fractions until the very end!



                      Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.



                      Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.






                      share|cite|improve this answer











                      $endgroup$









                      • 57




                        $begingroup$
                        I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 20:31






                      • 76




                        $begingroup$
                        @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 20:53








                      • 9




                        $begingroup$
                        I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 21:28








                      • 38




                        $begingroup$
                        @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 21:37






                      • 5




                        $begingroup$
                        In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                        $endgroup$
                        – André Nicolas
                        May 14 '14 at 15:50














                      495












                      495








                      495





                      $begingroup$

                      I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.



                      Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)



                      We want to solve the equation
                      $$ax^2+bx+c=0,$$
                      where $a ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.



                      Instead, multiply both sides by $4a$. We obtain the equivalent equation
                      $$4a^2x^2 +4abx+4ac=0.tag{1}$$
                      Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
                      $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
                      So our equation can be rewritten as
                      $$(2ax+b)^2 -b^2+4ac=0 tag{2}$$
                      or equivalently
                      $$(2ax+b)^2=b^2-4ac. tag{3}$$
                      Now it's all over. We find that
                      $$2ax+b=pmsqrt{b^2-4ac} tag{4}$$
                      and therefore
                      $$x=frac{-bpmsqrt{b^2-4ac}}{2a}. tag{5}$$

                      No fractions until the very end!



                      Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.



                      Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.






                      share|cite|improve this answer











                      $endgroup$



                      I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.



                      Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)



                      We want to solve the equation
                      $$ax^2+bx+c=0,$$
                      where $a ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.



                      Instead, multiply both sides by $4a$. We obtain the equivalent equation
                      $$4a^2x^2 +4abx+4ac=0.tag{1}$$
                      Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely,
                      $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$
                      So our equation can be rewritten as
                      $$(2ax+b)^2 -b^2+4ac=0 tag{2}$$
                      or equivalently
                      $$(2ax+b)^2=b^2-4ac. tag{3}$$
                      Now it's all over. We find that
                      $$2ax+b=pmsqrt{b^2-4ac} tag{4}$$
                      and therefore
                      $$x=frac{-bpmsqrt{b^2-4ac}}{2a}. tag{5}$$

                      No fractions until the very end!



                      Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.



                      Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Oct 22 '15 at 3:00









                      hlapointe

                      660721




                      660721










                      answered Jul 3 '11 at 16:53









                      André NicolasAndré Nicolas

                      452k36425810




                      452k36425810








                      • 57




                        $begingroup$
                        I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 20:31






                      • 76




                        $begingroup$
                        @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 20:53








                      • 9




                        $begingroup$
                        I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 21:28








                      • 38




                        $begingroup$
                        @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 21:37






                      • 5




                        $begingroup$
                        In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                        $endgroup$
                        – André Nicolas
                        May 14 '14 at 15:50














                      • 57




                        $begingroup$
                        I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 20:31






                      • 76




                        $begingroup$
                        @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 20:53








                      • 9




                        $begingroup$
                        I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 21:28








                      • 38




                        $begingroup$
                        @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 21:37






                      • 5




                        $begingroup$
                        In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                        $endgroup$
                        – André Nicolas
                        May 14 '14 at 15:50








                      57




                      57




                      $begingroup$
                      I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 20:31




                      $begingroup$
                      I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + frac{b}{a}x+ frac{c}{a}) = 0 iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = frac{b}{a}$, $C = frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind?
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 20:31




                      76




                      76




                      $begingroup$
                      @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 20:53






                      $begingroup$
                      @Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 20:53






                      9




                      9




                      $begingroup$
                      I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 21:28






                      $begingroup$
                      I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims.
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 21:28






                      38




                      38




                      $begingroup$
                      @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 21:37




                      $begingroup$
                      @Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 21:37




                      5




                      5




                      $begingroup$
                      In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                      $endgroup$
                      – André Nicolas
                      May 14 '14 at 15:50




                      $begingroup$
                      In modular arithmetic, the motivation is natural, since "fractions" lead to difficulties unless one takes an abstract point of view. But the argument is far more ancient than that. For a very long time, there was discomfort with algebraic manipulation of fractions. The procedure we have described delays that until the last step.
                      $endgroup$
                      – André Nicolas
                      May 14 '14 at 15:50











                      71












                      $begingroup$

                      Here is a slightly less ad-hoc approach to deriving the formula.



                      You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,times,div$ and $sqrt[n]{}$.



                      But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.



                      Intuitively because you know that $(r_1+r_2)=-frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=frac{-b}{2a}pmfrac{E}2$$



                      All we need to do then, is express $E=(r_1-r_2)$ using $+,-,times,div,sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,times,div$ and $a,b,c$ could possible be.



                      Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.



                      In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.



                      We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$



                      Evidently, now we have that $sqrt{D}=a(r_1-r_2)=aE$ and so $E=frac{sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=frac{-b}{2a}pmfrac{E}{2}$$ to $$r_1,r_2=frac{-b}{2a}pmfrac{sqrt{D}}{2a}$$ and finally $$r_1,r_2=frac{-bpmsqrt{b^2-4ac}}{2a}$$





                      The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.



                      What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of ${dfrac {p(a,b,c)}{q(a,b,c)}}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).



                      For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,times,div,sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                        $endgroup$
                        – Pedro Tamaroff
                        May 2 '12 at 23:03








                      • 1




                        $begingroup$
                        @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                        $endgroup$
                        – Vladimir Sotirov
                        Apr 19 '14 at 21:02






                      • 3




                        $begingroup$
                        @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                        $endgroup$
                        – Anastasiya-Romanova 秀
                        Apr 20 '14 at 8:59
















                      71












                      $begingroup$

                      Here is a slightly less ad-hoc approach to deriving the formula.



                      You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,times,div$ and $sqrt[n]{}$.



                      But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.



                      Intuitively because you know that $(r_1+r_2)=-frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=frac{-b}{2a}pmfrac{E}2$$



                      All we need to do then, is express $E=(r_1-r_2)$ using $+,-,times,div,sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,times,div$ and $a,b,c$ could possible be.



                      Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.



                      In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.



                      We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$



                      Evidently, now we have that $sqrt{D}=a(r_1-r_2)=aE$ and so $E=frac{sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=frac{-b}{2a}pmfrac{E}{2}$$ to $$r_1,r_2=frac{-b}{2a}pmfrac{sqrt{D}}{2a}$$ and finally $$r_1,r_2=frac{-bpmsqrt{b^2-4ac}}{2a}$$





                      The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.



                      What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of ${dfrac {p(a,b,c)}{q(a,b,c)}}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).



                      For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,times,div,sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                        $endgroup$
                        – Pedro Tamaroff
                        May 2 '12 at 23:03








                      • 1




                        $begingroup$
                        @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                        $endgroup$
                        – Vladimir Sotirov
                        Apr 19 '14 at 21:02






                      • 3




                        $begingroup$
                        @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                        $endgroup$
                        – Anastasiya-Romanova 秀
                        Apr 20 '14 at 8:59














                      71












                      71








                      71





                      $begingroup$

                      Here is a slightly less ad-hoc approach to deriving the formula.



                      You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,times,div$ and $sqrt[n]{}$.



                      But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.



                      Intuitively because you know that $(r_1+r_2)=-frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=frac{-b}{2a}pmfrac{E}2$$



                      All we need to do then, is express $E=(r_1-r_2)$ using $+,-,times,div,sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,times,div$ and $a,b,c$ could possible be.



                      Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.



                      In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.



                      We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$



                      Evidently, now we have that $sqrt{D}=a(r_1-r_2)=aE$ and so $E=frac{sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=frac{-b}{2a}pmfrac{E}{2}$$ to $$r_1,r_2=frac{-b}{2a}pmfrac{sqrt{D}}{2a}$$ and finally $$r_1,r_2=frac{-bpmsqrt{b^2-4ac}}{2a}$$





                      The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.



                      What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of ${dfrac {p(a,b,c)}{q(a,b,c)}}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).



                      For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,times,div,sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.






                      share|cite|improve this answer











                      $endgroup$



                      Here is a slightly less ad-hoc approach to deriving the formula.



                      You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,times,div$ and $sqrt[n]{}$.



                      But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.



                      Intuitively because you know that $(r_1+r_2)=-frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=frac{-b}{2a}pmfrac{E}2$$



                      All we need to do then, is express $E=(r_1-r_2)$ using $+,-,times,div,sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,times,div$ and $a,b,c$ could possible be.



                      Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.



                      In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.



                      We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$



                      Evidently, now we have that $sqrt{D}=a(r_1-r_2)=aE$ and so $E=frac{sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=frac{-b}{2a}pmfrac{E}{2}$$ to $$r_1,r_2=frac{-b}{2a}pmfrac{sqrt{D}}{2a}$$ and finally $$r_1,r_2=frac{-bpmsqrt{b^2-4ac}}{2a}$$





                      The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.



                      What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of ${dfrac {p(a,b,c)}{q(a,b,c)}}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).



                      For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,times,div,sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jul 26 '14 at 8:42









                      Hakim

                      9,15953157




                      9,15953157










                      answered Jul 3 '11 at 18:35









                      Vladimir SotirovVladimir Sotirov

                      8,63611948




                      8,63611948








                      • 1




                        $begingroup$
                        This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                        $endgroup$
                        – Pedro Tamaroff
                        May 2 '12 at 23:03








                      • 1




                        $begingroup$
                        @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                        $endgroup$
                        – Vladimir Sotirov
                        Apr 19 '14 at 21:02






                      • 3




                        $begingroup$
                        @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                        $endgroup$
                        – Anastasiya-Romanova 秀
                        Apr 20 '14 at 8:59














                      • 1




                        $begingroup$
                        This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                        $endgroup$
                        – Pedro Tamaroff
                        May 2 '12 at 23:03








                      • 1




                        $begingroup$
                        @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                        $endgroup$
                        – Vladimir Sotirov
                        Apr 19 '14 at 21:02






                      • 3




                        $begingroup$
                        @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                        $endgroup$
                        – Anastasiya-Romanova 秀
                        Apr 20 '14 at 8:59








                      1




                      1




                      $begingroup$
                      This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                      $endgroup$
                      – Pedro Tamaroff
                      May 2 '12 at 23:03






                      $begingroup$
                      This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1)
                      $endgroup$
                      – Pedro Tamaroff
                      May 2 '12 at 23:03






                      1




                      1




                      $begingroup$
                      @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                      $endgroup$
                      – Vladimir Sotirov
                      Apr 19 '14 at 21:02




                      $begingroup$
                      @V-Moy: the method is more complicated because instead of attempting to answer the simple question "How can we derive the quadratic formula?", it tries to answer the more complicated question "Why can we derive a quadratic formula?". The latter question is important because it turns out that while you can derive the cubic and quadric formulas, you cannot derive a quintic (or higher) formula. Galois theory (and the above way of thinking) arose precisely as an attempt to answer why you can for the former, and cannot for the latter.
                      $endgroup$
                      – Vladimir Sotirov
                      Apr 19 '14 at 21:02




                      3




                      3




                      $begingroup$
                      @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                      $endgroup$
                      – Anastasiya-Romanova 秀
                      Apr 20 '14 at 8:59




                      $begingroup$
                      @VladimirSotirov I know who Galois is but unfortunately I've never learnt his theory yet. Maybe you're right, but still it's hard for me to understand. I'm sorry for that, I'm just an 8th grader.
                      $endgroup$
                      – Anastasiya-Romanova 秀
                      Apr 20 '14 at 8:59











                      65












                      $begingroup$

                      Complete the square,



                      If you dare,



                      And the Answer



                      Will be there!






                      share|cite|improve this answer









                      $endgroup$









                      • 41




                        $begingroup$
                        What value does this add?
                        $endgroup$
                        – Peter Taylor
                        Jul 4 '11 at 12:29






                      • 33




                        $begingroup$
                        @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                        $endgroup$
                        – Mike Jones
                        Jul 4 '11 at 19:43
















                      65












                      $begingroup$

                      Complete the square,



                      If you dare,



                      And the Answer



                      Will be there!






                      share|cite|improve this answer









                      $endgroup$









                      • 41




                        $begingroup$
                        What value does this add?
                        $endgroup$
                        – Peter Taylor
                        Jul 4 '11 at 12:29






                      • 33




                        $begingroup$
                        @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                        $endgroup$
                        – Mike Jones
                        Jul 4 '11 at 19:43














                      65












                      65








                      65





                      $begingroup$

                      Complete the square,



                      If you dare,



                      And the Answer



                      Will be there!






                      share|cite|improve this answer









                      $endgroup$



                      Complete the square,



                      If you dare,



                      And the Answer



                      Will be there!







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jul 4 '11 at 0:05









                      ncmathsadistncmathsadist

                      42.7k260103




                      42.7k260103








                      • 41




                        $begingroup$
                        What value does this add?
                        $endgroup$
                        – Peter Taylor
                        Jul 4 '11 at 12:29






                      • 33




                        $begingroup$
                        @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                        $endgroup$
                        – Mike Jones
                        Jul 4 '11 at 19:43














                      • 41




                        $begingroup$
                        What value does this add?
                        $endgroup$
                        – Peter Taylor
                        Jul 4 '11 at 12:29






                      • 33




                        $begingroup$
                        @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                        $endgroup$
                        – Mike Jones
                        Jul 4 '11 at 19:43








                      41




                      41




                      $begingroup$
                      What value does this add?
                      $endgroup$
                      – Peter Taylor
                      Jul 4 '11 at 12:29




                      $begingroup$
                      What value does this add?
                      $endgroup$
                      – Peter Taylor
                      Jul 4 '11 at 12:29




                      33




                      33




                      $begingroup$
                      @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                      $endgroup$
                      – Mike Jones
                      Jul 4 '11 at 19:43




                      $begingroup$
                      @Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer.
                      $endgroup$
                      – Mike Jones
                      Jul 4 '11 at 19:43











                      56












                      $begingroup$

                      Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.



                      The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{bover a}x+{cover a}=0$. Now, that $bover a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{bover 2a})^2 = x^2+{bover a}x + {b^2over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $cover a$. A correction of $({cover a}-{b^2over 4a^2})$ will do this; we get $(x+{bover 2a})^2+({cover a}-{b^2over 4a^2}) = 0$.



                      But this is exactly what we wanted; we can move that second term over to the right and get $(x+{bover 2a})^2 = {b^2over 4a^2}-{cover a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4acover 4a^2}$ - just multiply the numerator and denominator of $cover a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{bover 2a} = pm sqrt{b^2-4acover 4a^2} = pm {sqrt{b^2-4ac}oversqrt{4a^2}} = {pmsqrt{b^2-4ac}over 2a}$. The last step is to subtract $bover 2a$ from both sides, finally giving the familiar: $x = {-bpmsqrt{b^2-4ac}over 2a}$






                      share|cite|improve this answer









                      $endgroup$









                      • 11




                        $begingroup$
                        Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 17:01










                      • $begingroup$
                        If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 17:51






                      • 3




                        $begingroup$
                        @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 18:03






                      • 1




                        $begingroup$
                        And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 18:07






                      • 3




                        $begingroup$
                        It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 22:04


















                      56












                      $begingroup$

                      Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.



                      The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{bover a}x+{cover a}=0$. Now, that $bover a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{bover 2a})^2 = x^2+{bover a}x + {b^2over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $cover a$. A correction of $({cover a}-{b^2over 4a^2})$ will do this; we get $(x+{bover 2a})^2+({cover a}-{b^2over 4a^2}) = 0$.



                      But this is exactly what we wanted; we can move that second term over to the right and get $(x+{bover 2a})^2 = {b^2over 4a^2}-{cover a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4acover 4a^2}$ - just multiply the numerator and denominator of $cover a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{bover 2a} = pm sqrt{b^2-4acover 4a^2} = pm {sqrt{b^2-4ac}oversqrt{4a^2}} = {pmsqrt{b^2-4ac}over 2a}$. The last step is to subtract $bover 2a$ from both sides, finally giving the familiar: $x = {-bpmsqrt{b^2-4ac}over 2a}$






                      share|cite|improve this answer









                      $endgroup$









                      • 11




                        $begingroup$
                        Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 17:01










                      • $begingroup$
                        If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 17:51






                      • 3




                        $begingroup$
                        @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 18:03






                      • 1




                        $begingroup$
                        And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 18:07






                      • 3




                        $begingroup$
                        It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 22:04
















                      56












                      56








                      56





                      $begingroup$

                      Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.



                      The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{bover a}x+{cover a}=0$. Now, that $bover a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{bover 2a})^2 = x^2+{bover a}x + {b^2over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $cover a$. A correction of $({cover a}-{b^2over 4a^2})$ will do this; we get $(x+{bover 2a})^2+({cover a}-{b^2over 4a^2}) = 0$.



                      But this is exactly what we wanted; we can move that second term over to the right and get $(x+{bover 2a})^2 = {b^2over 4a^2}-{cover a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4acover 4a^2}$ - just multiply the numerator and denominator of $cover a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{bover 2a} = pm sqrt{b^2-4acover 4a^2} = pm {sqrt{b^2-4ac}oversqrt{4a^2}} = {pmsqrt{b^2-4ac}over 2a}$. The last step is to subtract $bover 2a$ from both sides, finally giving the familiar: $x = {-bpmsqrt{b^2-4ac}over 2a}$






                      share|cite|improve this answer









                      $endgroup$



                      Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.



                      The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{bover a}x+{cover a}=0$. Now, that $bover a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{bover 2a})^2 = x^2+{bover a}x + {b^2over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $cover a$. A correction of $({cover a}-{b^2over 4a^2})$ will do this; we get $(x+{bover 2a})^2+({cover a}-{b^2over 4a^2}) = 0$.



                      But this is exactly what we wanted; we can move that second term over to the right and get $(x+{bover 2a})^2 = {b^2over 4a^2}-{cover a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4acover 4a^2}$ - just multiply the numerator and denominator of $cover a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{bover 2a} = pm sqrt{b^2-4acover 4a^2} = pm {sqrt{b^2-4ac}oversqrt{4a^2}} = {pmsqrt{b^2-4ac}over 2a}$. The last step is to subtract $bover 2a$ from both sides, finally giving the familiar: $x = {-bpmsqrt{b^2-4ac}over 2a}$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jul 3 '11 at 16:27









                      Steven StadnickiSteven Stadnicki

                      41.1k868122




                      41.1k868122








                      • 11




                        $begingroup$
                        Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 17:01










                      • $begingroup$
                        If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 17:51






                      • 3




                        $begingroup$
                        @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 18:03






                      • 1




                        $begingroup$
                        And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 18:07






                      • 3




                        $begingroup$
                        It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 22:04
















                      • 11




                        $begingroup$
                        Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 17:01










                      • $begingroup$
                        If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 17:51






                      • 3




                        $begingroup$
                        @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                        $endgroup$
                        – André Nicolas
                        Jul 3 '11 at 18:03






                      • 1




                        $begingroup$
                        And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                        $endgroup$
                        – dbasnett
                        Jul 3 '11 at 18:07






                      • 3




                        $begingroup$
                        It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                        $endgroup$
                        – Pete L. Clark
                        Jul 3 '11 at 22:04










                      11




                      11




                      $begingroup$
                      Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 17:01




                      $begingroup$
                      Almost at the end $sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $pm$ on top.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 17:01












                      $begingroup$
                      If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                      $endgroup$
                      – dbasnett
                      Jul 3 '11 at 17:51




                      $begingroup$
                      If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step?
                      $endgroup$
                      – dbasnett
                      Jul 3 '11 at 17:51




                      3




                      3




                      $begingroup$
                      @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 18:03




                      $begingroup$
                      @dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $sqrt{4a^2}=2|a|$.
                      $endgroup$
                      – André Nicolas
                      Jul 3 '11 at 18:03




                      1




                      1




                      $begingroup$
                      And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                      $endgroup$
                      – dbasnett
                      Jul 3 '11 at 18:07




                      $begingroup$
                      And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;)
                      $endgroup$
                      – dbasnett
                      Jul 3 '11 at 18:07




                      3




                      3




                      $begingroup$
                      It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 22:04






                      $begingroup$
                      It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = pm y$. Therefore in algebra whenever you extract a square root you have precisely a $pm$ ambiguity and should put a $pm$ in front of it. But we already have a $pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $sqrt{}$ always denotes the positive square root, $pm |a| = {a,-a} = pm a$.
                      $endgroup$
                      – Pete L. Clark
                      Jul 3 '11 at 22:04













                      50












                      $begingroup$

                      Proof without words.



                      completing the square



                      This one shows that $$ax^2+bx+c=aleft(x+dfrac b{2a}right)^2+c-dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.



                      Credits to LucasVB.



                      I hope this helps.

                      Best wishes, $mathcal H$akim.






                      share|cite|improve this answer









                      $endgroup$


















                        50












                        $begingroup$

                        Proof without words.



                        completing the square



                        This one shows that $$ax^2+bx+c=aleft(x+dfrac b{2a}right)^2+c-dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.



                        Credits to LucasVB.



                        I hope this helps.

                        Best wishes, $mathcal H$akim.






                        share|cite|improve this answer









                        $endgroup$
















                          50












                          50








                          50





                          $begingroup$

                          Proof without words.



                          completing the square



                          This one shows that $$ax^2+bx+c=aleft(x+dfrac b{2a}right)^2+c-dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.



                          Credits to LucasVB.



                          I hope this helps.

                          Best wishes, $mathcal H$akim.






                          share|cite|improve this answer









                          $endgroup$



                          Proof without words.



                          completing the square



                          This one shows that $$ax^2+bx+c=aleft(x+dfrac b{2a}right)^2+c-dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.



                          Credits to LucasVB.



                          I hope this helps.

                          Best wishes, $mathcal H$akim.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 13 '14 at 10:26









                          HakimHakim

                          9,15953157




                          9,15953157























                              40












                              $begingroup$

                              The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity
                              $$
                              aX^2+bX+c=aleft(X-frac{-b+sqrt{b^2-4ac}}{2a}right)left(X-frac{-b-sqrt{b^2-4ac}}{2a}right)
                              $$
                              holds for every $a$, $b$ and $c$.






                              share|cite|improve this answer









                              $endgroup$









                              • 1




                                $begingroup$
                                In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                                $endgroup$
                                – Martin Brandenburg
                                Jul 26 '14 at 9:05












                              • $begingroup$
                                Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                                $endgroup$
                                – Carsten S
                                Dec 10 '16 at 12:00
















                              40












                              $begingroup$

                              The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity
                              $$
                              aX^2+bX+c=aleft(X-frac{-b+sqrt{b^2-4ac}}{2a}right)left(X-frac{-b-sqrt{b^2-4ac}}{2a}right)
                              $$
                              holds for every $a$, $b$ and $c$.






                              share|cite|improve this answer









                              $endgroup$









                              • 1




                                $begingroup$
                                In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                                $endgroup$
                                – Martin Brandenburg
                                Jul 26 '14 at 9:05












                              • $begingroup$
                                Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                                $endgroup$
                                – Carsten S
                                Dec 10 '16 at 12:00














                              40












                              40








                              40





                              $begingroup$

                              The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity
                              $$
                              aX^2+bX+c=aleft(X-frac{-b+sqrt{b^2-4ac}}{2a}right)left(X-frac{-b-sqrt{b^2-4ac}}{2a}right)
                              $$
                              holds for every $a$, $b$ and $c$.






                              share|cite|improve this answer









                              $endgroup$



                              The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity
                              $$
                              aX^2+bX+c=aleft(X-frac{-b+sqrt{b^2-4ac}}{2a}right)left(X-frac{-b-sqrt{b^2-4ac}}{2a}right)
                              $$
                              holds for every $a$, $b$ and $c$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jul 3 '11 at 16:45









                              Andrea MoriAndrea Mori

                              19.5k13466




                              19.5k13466








                              • 1




                                $begingroup$
                                In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                                $endgroup$
                                – Martin Brandenburg
                                Jul 26 '14 at 9:05












                              • $begingroup$
                                Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                                $endgroup$
                                – Carsten S
                                Dec 10 '16 at 12:00














                              • 1




                                $begingroup$
                                In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                                $endgroup$
                                – Martin Brandenburg
                                Jul 26 '14 at 9:05












                              • $begingroup$
                                Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                                $endgroup$
                                – Carsten S
                                Dec 10 '16 at 12:00








                              1




                              1




                              $begingroup$
                              In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                              $endgroup$
                              – Martin Brandenburg
                              Jul 26 '14 at 9:05






                              $begingroup$
                              In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$.
                              $endgroup$
                              – Martin Brandenburg
                              Jul 26 '14 at 9:05














                              $begingroup$
                              Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                              $endgroup$
                              – Carsten S
                              Dec 10 '16 at 12:00




                              $begingroup$
                              Yes. However, it is not immediately clear that this something about the roots of the polynomial in the case $4ac>b^2$.
                              $endgroup$
                              – Carsten S
                              Dec 10 '16 at 12:00











                              39












                              $begingroup$

                              Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.



                              Try working backwards!



                              Assuming that the quadratic formula holds true,$$x = {-b pmsqrt{b^2 - 4ac} over 2a} $$Isolate $x$.$$begin{align}2ax &=& -b pm sqrt{b^2 - 4ac} \ 2ax + b & = &pmsqrt{b^2 - 4ac} \ (2ax + b)^2 & = & b^2 - 4ac \ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \ 4a^2x^2 + 4abx + 4ac & = & 0 \ ax^2 + bx + c & = & 0 end{align}$$
                              Note that we divided both sides by $4a$ in the last step assuming a $ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}cdots a_0x^0$, $a_n ne 0$.



                              Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                                $endgroup$
                                – Deepak Gupta
                                Sep 18 '15 at 7:20
















                              39












                              $begingroup$

                              Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.



                              Try working backwards!



                              Assuming that the quadratic formula holds true,$$x = {-b pmsqrt{b^2 - 4ac} over 2a} $$Isolate $x$.$$begin{align}2ax &=& -b pm sqrt{b^2 - 4ac} \ 2ax + b & = &pmsqrt{b^2 - 4ac} \ (2ax + b)^2 & = & b^2 - 4ac \ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \ 4a^2x^2 + 4abx + 4ac & = & 0 \ ax^2 + bx + c & = & 0 end{align}$$
                              Note that we divided both sides by $4a$ in the last step assuming a $ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}cdots a_0x^0$, $a_n ne 0$.



                              Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                                $endgroup$
                                – Deepak Gupta
                                Sep 18 '15 at 7:20














                              39












                              39








                              39





                              $begingroup$

                              Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.



                              Try working backwards!



                              Assuming that the quadratic formula holds true,$$x = {-b pmsqrt{b^2 - 4ac} over 2a} $$Isolate $x$.$$begin{align}2ax &=& -b pm sqrt{b^2 - 4ac} \ 2ax + b & = &pmsqrt{b^2 - 4ac} \ (2ax + b)^2 & = & b^2 - 4ac \ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \ 4a^2x^2 + 4abx + 4ac & = & 0 \ ax^2 + bx + c & = & 0 end{align}$$
                              Note that we divided both sides by $4a$ in the last step assuming a $ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}cdots a_0x^0$, $a_n ne 0$.



                              Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!






                              share|cite|improve this answer











                              $endgroup$



                              Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.



                              Try working backwards!



                              Assuming that the quadratic formula holds true,$$x = {-b pmsqrt{b^2 - 4ac} over 2a} $$Isolate $x$.$$begin{align}2ax &=& -b pm sqrt{b^2 - 4ac} \ 2ax + b & = &pmsqrt{b^2 - 4ac} \ (2ax + b)^2 & = & b^2 - 4ac \ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \ 4a^2x^2 + 4abx + 4ac & = & 0 \ ax^2 + bx + c & = & 0 end{align}$$
                              Note that we divided both sides by $4a$ in the last step assuming a $ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}cdots a_0x^0$, $a_n ne 0$.



                              Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Apr 13 '17 at 12:20









                              Community

                              1




                              1










                              answered Sep 25 '12 at 14:02









                              Parth KohliParth Kohli

                              6,01512860




                              6,01512860












                              • $begingroup$
                                From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                                $endgroup$
                                – Deepak Gupta
                                Sep 18 '15 at 7:20


















                              • $begingroup$
                                From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                                $endgroup$
                                – Deepak Gupta
                                Sep 18 '15 at 7:20
















                              $begingroup$
                              From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                              $endgroup$
                              – Deepak Gupta
                              Sep 18 '15 at 7:20




                              $begingroup$
                              From strict logic standpoint, if $p implies q$ then, not necessarily, $q implies p$ (except under certain conditions, which, incidentally happen to be true in this case, but nonetheless must be explicitly stated). $p = $ "If $x = alpha$ or $beta$ then $ax^2+bx+c=0$" and $q=$ "If $ax^2+bx+c=0$ then $x=alpha$ or $beta$"
                              $endgroup$
                              – Deepak Gupta
                              Sep 18 '15 at 7:20











                              28












                              $begingroup$

                              First, let's examine an analogous simpler special case. Why does the difference of squares formula $rm: x^2 - a^2 = (x-a) (x+a):$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $rm (x-a) (x+a) = x^2 - a x + x a - a^2 $ which indeed equals $rm x^2 - a^2 $ as long as $rm a x = x a $ for all $rm:x:,:$ i.e. as long as $rm:a:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.



                              In fact, if we consider both $rm:x:$ and $rm:a:$ as indeterminates, then we can specialize the "generic" factorization $rm: x^2 - a^2 = (x-a) (x + a):$ in the ring $rm:mathbb Z[x,a]:$ to any commutative ring by using an evaluation homomorphism mapping $rm:x,a:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.



                              A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $rm:2:a:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $rm:e = 1/(2a), d^2 = b^2 - 4ac:,:$ so the proofs works naturally in the ring $rm:mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac):.:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.



                              Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $rm det (1+AB)=det (1+BA) $ that proceeds by universally cancelling $rm det A $ from the $rm det $ of $rm (1+A B) A = A (1+B A),$ in $,rmBbb Z[a_{,ij},b_{,ij}],$ where the matrix entries are indeterminates $rm,a_{,ij},b_{,ij}.,$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).



                              REMARK $ $ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $rm:x^2 = 1:$ has roots $rm:pm1,:pm3 in mathbb Z/8:,:$ the ring of integers modulo $8:.:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $rm xy = 0 Rightarrow x=0 or y=0:.;$ Indeed, a polynomial $rm f(x)in D[x] $ has at most $rm deg f $ roots in the ring $rm:D $ iff $rm D:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $rm mathbb Z/m $ by showing that, given any $rm:f(x):$ with more roots than its degree, we can quickly compute a nontrivial factor of $rm:m:$ via a $rm:gcd:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $rm:m:$ by searching for a nontrivial square root in $rm: mathbb Z/m:,:$ e.g. a square root of $1$ that is not $:pm 1$.






                              share|cite|improve this answer











                              $endgroup$









                              • 4




                                $begingroup$
                                Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:30






                              • 3




                                $begingroup$
                                @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:37






                              • 3




                                $begingroup$
                                @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:42








                              • 3




                                $begingroup$
                                $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:56






                              • 12




                                $begingroup$
                                @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                                $endgroup$
                                – Bill Dubuque
                                Jul 4 '11 at 14:05
















                              28












                              $begingroup$

                              First, let's examine an analogous simpler special case. Why does the difference of squares formula $rm: x^2 - a^2 = (x-a) (x+a):$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $rm (x-a) (x+a) = x^2 - a x + x a - a^2 $ which indeed equals $rm x^2 - a^2 $ as long as $rm a x = x a $ for all $rm:x:,:$ i.e. as long as $rm:a:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.



                              In fact, if we consider both $rm:x:$ and $rm:a:$ as indeterminates, then we can specialize the "generic" factorization $rm: x^2 - a^2 = (x-a) (x + a):$ in the ring $rm:mathbb Z[x,a]:$ to any commutative ring by using an evaluation homomorphism mapping $rm:x,a:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.



                              A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $rm:2:a:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $rm:e = 1/(2a), d^2 = b^2 - 4ac:,:$ so the proofs works naturally in the ring $rm:mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac):.:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.



                              Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $rm det (1+AB)=det (1+BA) $ that proceeds by universally cancelling $rm det A $ from the $rm det $ of $rm (1+A B) A = A (1+B A),$ in $,rmBbb Z[a_{,ij},b_{,ij}],$ where the matrix entries are indeterminates $rm,a_{,ij},b_{,ij}.,$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).



                              REMARK $ $ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $rm:x^2 = 1:$ has roots $rm:pm1,:pm3 in mathbb Z/8:,:$ the ring of integers modulo $8:.:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $rm xy = 0 Rightarrow x=0 or y=0:.;$ Indeed, a polynomial $rm f(x)in D[x] $ has at most $rm deg f $ roots in the ring $rm:D $ iff $rm D:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $rm mathbb Z/m $ by showing that, given any $rm:f(x):$ with more roots than its degree, we can quickly compute a nontrivial factor of $rm:m:$ via a $rm:gcd:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $rm:m:$ by searching for a nontrivial square root in $rm: mathbb Z/m:,:$ e.g. a square root of $1$ that is not $:pm 1$.






                              share|cite|improve this answer











                              $endgroup$









                              • 4




                                $begingroup$
                                Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:30






                              • 3




                                $begingroup$
                                @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:37






                              • 3




                                $begingroup$
                                @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:42








                              • 3




                                $begingroup$
                                $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:56






                              • 12




                                $begingroup$
                                @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                                $endgroup$
                                – Bill Dubuque
                                Jul 4 '11 at 14:05














                              28












                              28








                              28





                              $begingroup$

                              First, let's examine an analogous simpler special case. Why does the difference of squares formula $rm: x^2 - a^2 = (x-a) (x+a):$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $rm (x-a) (x+a) = x^2 - a x + x a - a^2 $ which indeed equals $rm x^2 - a^2 $ as long as $rm a x = x a $ for all $rm:x:,:$ i.e. as long as $rm:a:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.



                              In fact, if we consider both $rm:x:$ and $rm:a:$ as indeterminates, then we can specialize the "generic" factorization $rm: x^2 - a^2 = (x-a) (x + a):$ in the ring $rm:mathbb Z[x,a]:$ to any commutative ring by using an evaluation homomorphism mapping $rm:x,a:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.



                              A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $rm:2:a:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $rm:e = 1/(2a), d^2 = b^2 - 4ac:,:$ so the proofs works naturally in the ring $rm:mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac):.:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.



                              Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $rm det (1+AB)=det (1+BA) $ that proceeds by universally cancelling $rm det A $ from the $rm det $ of $rm (1+A B) A = A (1+B A),$ in $,rmBbb Z[a_{,ij},b_{,ij}],$ where the matrix entries are indeterminates $rm,a_{,ij},b_{,ij}.,$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).



                              REMARK $ $ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $rm:x^2 = 1:$ has roots $rm:pm1,:pm3 in mathbb Z/8:,:$ the ring of integers modulo $8:.:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $rm xy = 0 Rightarrow x=0 or y=0:.;$ Indeed, a polynomial $rm f(x)in D[x] $ has at most $rm deg f $ roots in the ring $rm:D $ iff $rm D:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $rm mathbb Z/m $ by showing that, given any $rm:f(x):$ with more roots than its degree, we can quickly compute a nontrivial factor of $rm:m:$ via a $rm:gcd:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $rm:m:$ by searching for a nontrivial square root in $rm: mathbb Z/m:,:$ e.g. a square root of $1$ that is not $:pm 1$.






                              share|cite|improve this answer











                              $endgroup$



                              First, let's examine an analogous simpler special case. Why does the difference of squares formula $rm: x^2 - a^2 = (x-a) (x+a):$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $rm (x-a) (x+a) = x^2 - a x + x a - a^2 $ which indeed equals $rm x^2 - a^2 $ as long as $rm a x = x a $ for all $rm:x:,:$ i.e. as long as $rm:a:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.



                              In fact, if we consider both $rm:x:$ and $rm:a:$ as indeterminates, then we can specialize the "generic" factorization $rm: x^2 - a^2 = (x-a) (x + a):$ in the ring $rm:mathbb Z[x,a]:$ to any commutative ring by using an evaluation homomorphism mapping $rm:x,a:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.



                              A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $rm:2:a:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $rm:e = 1/(2a), d^2 = b^2 - 4ac:,:$ so the proofs works naturally in the ring $rm:mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac):.:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.



                              Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $rm det (1+AB)=det (1+BA) $ that proceeds by universally cancelling $rm det A $ from the $rm det $ of $rm (1+A B) A = A (1+B A),$ in $,rmBbb Z[a_{,ij},b_{,ij}],$ where the matrix entries are indeterminates $rm,a_{,ij},b_{,ij}.,$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).



                              REMARK $ $ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $rm:x^2 = 1:$ has roots $rm:pm1,:pm3 in mathbb Z/8:,:$ the ring of integers modulo $8:.:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $rm xy = 0 Rightarrow x=0 or y=0:.;$ Indeed, a polynomial $rm f(x)in D[x] $ has at most $rm deg f $ roots in the ring $rm:D $ iff $rm D:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $rm mathbb Z/m $ by showing that, given any $rm:f(x):$ with more roots than its degree, we can quickly compute a nontrivial factor of $rm:m:$ via a $rm:gcd:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $rm:m:$ by searching for a nontrivial square root in $rm: mathbb Z/m:,:$ e.g. a square root of $1$ that is not $:pm 1$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Sep 21 '17 at 17:05

























                              answered Jul 3 '11 at 22:30









                              Bill DubuqueBill Dubuque

                              210k29192640




                              210k29192640








                              • 4




                                $begingroup$
                                Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:30






                              • 3




                                $begingroup$
                                @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:37






                              • 3




                                $begingroup$
                                @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:42








                              • 3




                                $begingroup$
                                $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:56






                              • 12




                                $begingroup$
                                @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                                $endgroup$
                                – Bill Dubuque
                                Jul 4 '11 at 14:05














                              • 4




                                $begingroup$
                                Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:30






                              • 3




                                $begingroup$
                                @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:37






                              • 3




                                $begingroup$
                                @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                                $endgroup$
                                – Bill Dubuque
                                Jul 3 '11 at 23:42








                              • 3




                                $begingroup$
                                $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                                $endgroup$
                                – Pete L. Clark
                                Jul 3 '11 at 23:56






                              • 12




                                $begingroup$
                                @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                                $endgroup$
                                – Bill Dubuque
                                Jul 4 '11 at 14:05








                              4




                              4




                              $begingroup$
                              Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                              $endgroup$
                              – Pete L. Clark
                              Jul 3 '11 at 23:30




                              $begingroup$
                              Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way.
                              $endgroup$
                              – Pete L. Clark
                              Jul 3 '11 at 23:30




                              3




                              3




                              $begingroup$
                              @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                              $endgroup$
                              – Bill Dubuque
                              Jul 3 '11 at 23:37




                              $begingroup$
                              @Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $sqrt{x}$ or $pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above.
                              $endgroup$
                              – Bill Dubuque
                              Jul 3 '11 at 23:37




                              3




                              3




                              $begingroup$
                              @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                              $endgroup$
                              – Bill Dubuque
                              Jul 3 '11 at 23:42






                              $begingroup$
                              @Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything.
                              $endgroup$
                              – Bill Dubuque
                              Jul 3 '11 at 23:42






                              3




                              3




                              $begingroup$
                              $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                              $endgroup$
                              – Pete L. Clark
                              Jul 3 '11 at 23:56




                              $begingroup$
                              $@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments".
                              $endgroup$
                              – Pete L. Clark
                              Jul 3 '11 at 23:56




                              12




                              12




                              $begingroup$
                              @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                              $endgroup$
                              – Bill Dubuque
                              Jul 4 '11 at 14:05




                              $begingroup$
                              @Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns.
                              $endgroup$
                              – Bill Dubuque
                              Jul 4 '11 at 14:05











                              18












                              $begingroup$

                              The formula comes from the technique known as completing the square.






                              share|cite|improve this answer









                              $endgroup$


















                                18












                                $begingroup$

                                The formula comes from the technique known as completing the square.






                                share|cite|improve this answer









                                $endgroup$
















                                  18












                                  18








                                  18





                                  $begingroup$

                                  The formula comes from the technique known as completing the square.






                                  share|cite|improve this answer









                                  $endgroup$



                                  The formula comes from the technique known as completing the square.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Jul 3 '11 at 16:16









                                  lhflhf

                                  164k10170395




                                  164k10170395























                                      18












                                      $begingroup$

                                      I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.




                                      Consider any quadratic equation



                                      $$ax^{2}+bx+c=0,qquad (aneq 0).qquad (1)$$



                                      Multiplying both sides by $1/a$ we get
                                      the equivalent equation



                                      $$x^{2}+frac{b}{a}x+frac{c}{a}=0.$$



                                      We are going to show that it is possible to find $h$ and $alpha $ such that



                                      $$x^{2}+frac{b}{a}x+frac{c}{a}=(x+h)^{2}-alpha .$$



                                      Expanding the RHS gives



                                      $$x^{2}+frac{b}{a}x+frac{c}{a}=x^{2}+2hx+h^{2}-alpha .$$



                                      This means, applying the method of undetermined coefficients, that



                                      $$left{ begin{array}{l} frac{b}{a}=2h \ frac{c}{a}=h^{2}-alpha. end{array} right. $$



                                      Hence



                                      $$left{ begin{array}{l}h=frac{b}{2a} \ alpha=h^{2}-frac{c}{a}=frac{b^{2}}{4a^{2}}-frac{c}{a}=frac{b^{2}-4ac}{4a^{2}}.end{array}right. qquad (2)$$



                                      In this way the given equation is reduced to the binomial equation in
                                      $x+h$



                                      $$left( x+hright) ^{2}-alpha =0qquad text{equivalent to}qquad left( x+hright) ^{2}=alpha $$



                                      and thus it is satisfied when



                                      $$x+h=sqrt{alpha }qquad text{or}qquad x+h=-sqrt{alpha },$$



                                      i.e. when



                                      $$x=-h+sqrt{alpha }qquad text{or}qquad x=-h-sqrt{alpha }.$$



                                      Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $alpha $ by their expressions given by $(2)$, yields



                                      $$x_{1}=frac{-b+sqrt{b^{2}-4ac}}{2a}qquad text{or}qquad x_{2}=frac{-b-sqrt{b^{2}-4ac}}{2a}.qquad (3)$$




                                      As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.






                                      share|cite|improve this answer











                                      $endgroup$


















                                        18












                                        $begingroup$

                                        I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.




                                        Consider any quadratic equation



                                        $$ax^{2}+bx+c=0,qquad (aneq 0).qquad (1)$$



                                        Multiplying both sides by $1/a$ we get
                                        the equivalent equation



                                        $$x^{2}+frac{b}{a}x+frac{c}{a}=0.$$



                                        We are going to show that it is possible to find $h$ and $alpha $ such that



                                        $$x^{2}+frac{b}{a}x+frac{c}{a}=(x+h)^{2}-alpha .$$



                                        Expanding the RHS gives



                                        $$x^{2}+frac{b}{a}x+frac{c}{a}=x^{2}+2hx+h^{2}-alpha .$$



                                        This means, applying the method of undetermined coefficients, that



                                        $$left{ begin{array}{l} frac{b}{a}=2h \ frac{c}{a}=h^{2}-alpha. end{array} right. $$



                                        Hence



                                        $$left{ begin{array}{l}h=frac{b}{2a} \ alpha=h^{2}-frac{c}{a}=frac{b^{2}}{4a^{2}}-frac{c}{a}=frac{b^{2}-4ac}{4a^{2}}.end{array}right. qquad (2)$$



                                        In this way the given equation is reduced to the binomial equation in
                                        $x+h$



                                        $$left( x+hright) ^{2}-alpha =0qquad text{equivalent to}qquad left( x+hright) ^{2}=alpha $$



                                        and thus it is satisfied when



                                        $$x+h=sqrt{alpha }qquad text{or}qquad x+h=-sqrt{alpha },$$



                                        i.e. when



                                        $$x=-h+sqrt{alpha }qquad text{or}qquad x=-h-sqrt{alpha }.$$



                                        Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $alpha $ by their expressions given by $(2)$, yields



                                        $$x_{1}=frac{-b+sqrt{b^{2}-4ac}}{2a}qquad text{or}qquad x_{2}=frac{-b-sqrt{b^{2}-4ac}}{2a}.qquad (3)$$




                                        As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.






                                        share|cite|improve this answer











                                        $endgroup$
















                                          18












                                          18








                                          18





                                          $begingroup$

                                          I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.




                                          Consider any quadratic equation



                                          $$ax^{2}+bx+c=0,qquad (aneq 0).qquad (1)$$



                                          Multiplying both sides by $1/a$ we get
                                          the equivalent equation



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=0.$$



                                          We are going to show that it is possible to find $h$ and $alpha $ such that



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=(x+h)^{2}-alpha .$$



                                          Expanding the RHS gives



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=x^{2}+2hx+h^{2}-alpha .$$



                                          This means, applying the method of undetermined coefficients, that



                                          $$left{ begin{array}{l} frac{b}{a}=2h \ frac{c}{a}=h^{2}-alpha. end{array} right. $$



                                          Hence



                                          $$left{ begin{array}{l}h=frac{b}{2a} \ alpha=h^{2}-frac{c}{a}=frac{b^{2}}{4a^{2}}-frac{c}{a}=frac{b^{2}-4ac}{4a^{2}}.end{array}right. qquad (2)$$



                                          In this way the given equation is reduced to the binomial equation in
                                          $x+h$



                                          $$left( x+hright) ^{2}-alpha =0qquad text{equivalent to}qquad left( x+hright) ^{2}=alpha $$



                                          and thus it is satisfied when



                                          $$x+h=sqrt{alpha }qquad text{or}qquad x+h=-sqrt{alpha },$$



                                          i.e. when



                                          $$x=-h+sqrt{alpha }qquad text{or}qquad x=-h-sqrt{alpha }.$$



                                          Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $alpha $ by their expressions given by $(2)$, yields



                                          $$x_{1}=frac{-b+sqrt{b^{2}-4ac}}{2a}qquad text{or}qquad x_{2}=frac{-b-sqrt{b^{2}-4ac}}{2a}.qquad (3)$$




                                          As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.






                                          share|cite|improve this answer











                                          $endgroup$



                                          I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.




                                          Consider any quadratic equation



                                          $$ax^{2}+bx+c=0,qquad (aneq 0).qquad (1)$$



                                          Multiplying both sides by $1/a$ we get
                                          the equivalent equation



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=0.$$



                                          We are going to show that it is possible to find $h$ and $alpha $ such that



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=(x+h)^{2}-alpha .$$



                                          Expanding the RHS gives



                                          $$x^{2}+frac{b}{a}x+frac{c}{a}=x^{2}+2hx+h^{2}-alpha .$$



                                          This means, applying the method of undetermined coefficients, that



                                          $$left{ begin{array}{l} frac{b}{a}=2h \ frac{c}{a}=h^{2}-alpha. end{array} right. $$



                                          Hence



                                          $$left{ begin{array}{l}h=frac{b}{2a} \ alpha=h^{2}-frac{c}{a}=frac{b^{2}}{4a^{2}}-frac{c}{a}=frac{b^{2}-4ac}{4a^{2}}.end{array}right. qquad (2)$$



                                          In this way the given equation is reduced to the binomial equation in
                                          $x+h$



                                          $$left( x+hright) ^{2}-alpha =0qquad text{equivalent to}qquad left( x+hright) ^{2}=alpha $$



                                          and thus it is satisfied when



                                          $$x+h=sqrt{alpha }qquad text{or}qquad x+h=-sqrt{alpha },$$



                                          i.e. when



                                          $$x=-h+sqrt{alpha }qquad text{or}qquad x=-h-sqrt{alpha }.$$



                                          Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $alpha $ by their expressions given by $(2)$, yields



                                          $$x_{1}=frac{-b+sqrt{b^{2}-4ac}}{2a}qquad text{or}qquad x_{2}=frac{-b-sqrt{b^{2}-4ac}}{2a}.qquad (3)$$




                                          As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          answered Jul 3 '11 at 23:40


























                                          community wiki





                                          Américo Tavares
























                                              11












                                              $begingroup$

                                              Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).

                                              So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)






                                              share|cite|improve this answer









                                              $endgroup$









                                              • 9




                                                $begingroup$
                                                I thought that was implied by "degree two"!
                                                $endgroup$
                                                – The Chaz 2.0
                                                Jul 4 '11 at 5:40
















                                              11












                                              $begingroup$

                                              Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).

                                              So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)






                                              share|cite|improve this answer









                                              $endgroup$









                                              • 9




                                                $begingroup$
                                                I thought that was implied by "degree two"!
                                                $endgroup$
                                                – The Chaz 2.0
                                                Jul 4 '11 at 5:40














                                              11












                                              11








                                              11





                                              $begingroup$

                                              Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).

                                              So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)






                                              share|cite|improve this answer









                                              $endgroup$



                                              Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).

                                              So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Jul 3 '11 at 16:33









                                              The Chaz 2.0The Chaz 2.0

                                              8,11363575




                                              8,11363575








                                              • 9




                                                $begingroup$
                                                I thought that was implied by "degree two"!
                                                $endgroup$
                                                – The Chaz 2.0
                                                Jul 4 '11 at 5:40














                                              • 9




                                                $begingroup$
                                                I thought that was implied by "degree two"!
                                                $endgroup$
                                                – The Chaz 2.0
                                                Jul 4 '11 at 5:40








                                              9




                                              9




                                              $begingroup$
                                              I thought that was implied by "degree two"!
                                              $endgroup$
                                              – The Chaz 2.0
                                              Jul 4 '11 at 5:40




                                              $begingroup$
                                              I thought that was implied by "degree two"!
                                              $endgroup$
                                              – The Chaz 2.0
                                              Jul 4 '11 at 5:40











                                              11












                                              $begingroup$

                                              This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $mathbb{R}$ or $mathbb{Q}$).



                                              Step 1. The standard identity
                                              begin{equation}
                                              (x + y)^2 = (x-y)^2 + 4xy qquad qquad (1)
                                              end{equation}



                                              Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = frac{1}{2}(x+y) + (x-y)$ and $y = frac{1}{2}(x+y) + (x-y)$.



                                              Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a not= 0$. Dividing by $a$ yields $x^2 + frac{b}{a}x +frac{c}{a} = 0$ or $x(x + frac{b}{a}) = -frac{c}{a}$. Setting $y = x + frac{b}{a}$, we get $xy= -frac{c}{a}$ and $x - y = -frac{b}{a}$.
                                              The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.



                                              I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.






                                              share|cite|improve this answer









                                              $endgroup$


















                                                11












                                                $begingroup$

                                                This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $mathbb{R}$ or $mathbb{Q}$).



                                                Step 1. The standard identity
                                                begin{equation}
                                                (x + y)^2 = (x-y)^2 + 4xy qquad qquad (1)
                                                end{equation}



                                                Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = frac{1}{2}(x+y) + (x-y)$ and $y = frac{1}{2}(x+y) + (x-y)$.



                                                Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a not= 0$. Dividing by $a$ yields $x^2 + frac{b}{a}x +frac{c}{a} = 0$ or $x(x + frac{b}{a}) = -frac{c}{a}$. Setting $y = x + frac{b}{a}$, we get $xy= -frac{c}{a}$ and $x - y = -frac{b}{a}$.
                                                The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.



                                                I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.






                                                share|cite|improve this answer









                                                $endgroup$
















                                                  11












                                                  11








                                                  11





                                                  $begingroup$

                                                  This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $mathbb{R}$ or $mathbb{Q}$).



                                                  Step 1. The standard identity
                                                  begin{equation}
                                                  (x + y)^2 = (x-y)^2 + 4xy qquad qquad (1)
                                                  end{equation}



                                                  Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = frac{1}{2}(x+y) + (x-y)$ and $y = frac{1}{2}(x+y) + (x-y)$.



                                                  Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a not= 0$. Dividing by $a$ yields $x^2 + frac{b}{a}x +frac{c}{a} = 0$ or $x(x + frac{b}{a}) = -frac{c}{a}$. Setting $y = x + frac{b}{a}$, we get $xy= -frac{c}{a}$ and $x - y = -frac{b}{a}$.
                                                  The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.



                                                  I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.






                                                  share|cite|improve this answer









                                                  $endgroup$



                                                  This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $mathbb{R}$ or $mathbb{Q}$).



                                                  Step 1. The standard identity
                                                  begin{equation}
                                                  (x + y)^2 = (x-y)^2 + 4xy qquad qquad (1)
                                                  end{equation}



                                                  Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = frac{1}{2}(x+y) + (x-y)$ and $y = frac{1}{2}(x+y) + (x-y)$.



                                                  Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a not= 0$. Dividing by $a$ yields $x^2 + frac{b}{a}x +frac{c}{a} = 0$ or $x(x + frac{b}{a}) = -frac{c}{a}$. Setting $y = x + frac{b}{a}$, we get $xy= -frac{c}{a}$ and $x - y = -frac{b}{a}$.
                                                  The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.



                                                  I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered Sep 30 '13 at 21:38









                                                  J.-E. PinJ.-E. Pin

                                                  18.4k21754




                                                  18.4k21754























                                                      9












                                                      $begingroup$

                                                      You could also prove it using Tschirnhaus Transformation.



                                                      Let $,x=y-frac{b}{2a}$.$,$ Then $,ax^2+bx+c=0,$ becomes
                                                      $$begin{align}ay^2-frac{b^2}{4a}+c&=0\
                                                      iff y^2&=frac{b^2-4ac}{4a^2}\
                                                      iff y&=frac{pmsqrt{b^2-4ac}}{2a}\
                                                      iff x&=frac{pmsqrt{b^2-4ac}}{2a}-frac{b}{2a}\
                                                      iff x&=frac{-bpmsqrt{b^2-4ac}}{2a}end{align}$$





                                                      In general, the substitution for $, a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0,$ is $,x=y-frac{a_{n-1}}{na_n}$,



                                                      which leaves the polynomial without a term of degree $n-1$.






                                                      share|cite|improve this answer











                                                      $endgroup$













                                                      • $begingroup$
                                                        Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                        $endgroup$
                                                        – Daniel W. Farlow
                                                        Feb 8 '15 at 4:19






                                                      • 1




                                                        $begingroup$
                                                        @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                        $endgroup$
                                                        – user26486
                                                        Feb 8 '15 at 4:48








                                                      • 3




                                                        $begingroup$
                                                        I agree with @induktio
                                                        $endgroup$
                                                        – fancynancy
                                                        Feb 8 '15 at 5:28






                                                      • 3




                                                        $begingroup$
                                                        I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                        $endgroup$
                                                        – user26486
                                                        Feb 19 '15 at 15:01
















                                                      9












                                                      $begingroup$

                                                      You could also prove it using Tschirnhaus Transformation.



                                                      Let $,x=y-frac{b}{2a}$.$,$ Then $,ax^2+bx+c=0,$ becomes
                                                      $$begin{align}ay^2-frac{b^2}{4a}+c&=0\
                                                      iff y^2&=frac{b^2-4ac}{4a^2}\
                                                      iff y&=frac{pmsqrt{b^2-4ac}}{2a}\
                                                      iff x&=frac{pmsqrt{b^2-4ac}}{2a}-frac{b}{2a}\
                                                      iff x&=frac{-bpmsqrt{b^2-4ac}}{2a}end{align}$$





                                                      In general, the substitution for $, a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0,$ is $,x=y-frac{a_{n-1}}{na_n}$,



                                                      which leaves the polynomial without a term of degree $n-1$.






                                                      share|cite|improve this answer











                                                      $endgroup$













                                                      • $begingroup$
                                                        Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                        $endgroup$
                                                        – Daniel W. Farlow
                                                        Feb 8 '15 at 4:19






                                                      • 1




                                                        $begingroup$
                                                        @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                        $endgroup$
                                                        – user26486
                                                        Feb 8 '15 at 4:48








                                                      • 3




                                                        $begingroup$
                                                        I agree with @induktio
                                                        $endgroup$
                                                        – fancynancy
                                                        Feb 8 '15 at 5:28






                                                      • 3




                                                        $begingroup$
                                                        I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                        $endgroup$
                                                        – user26486
                                                        Feb 19 '15 at 15:01














                                                      9












                                                      9








                                                      9





                                                      $begingroup$

                                                      You could also prove it using Tschirnhaus Transformation.



                                                      Let $,x=y-frac{b}{2a}$.$,$ Then $,ax^2+bx+c=0,$ becomes
                                                      $$begin{align}ay^2-frac{b^2}{4a}+c&=0\
                                                      iff y^2&=frac{b^2-4ac}{4a^2}\
                                                      iff y&=frac{pmsqrt{b^2-4ac}}{2a}\
                                                      iff x&=frac{pmsqrt{b^2-4ac}}{2a}-frac{b}{2a}\
                                                      iff x&=frac{-bpmsqrt{b^2-4ac}}{2a}end{align}$$





                                                      In general, the substitution for $, a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0,$ is $,x=y-frac{a_{n-1}}{na_n}$,



                                                      which leaves the polynomial without a term of degree $n-1$.






                                                      share|cite|improve this answer











                                                      $endgroup$



                                                      You could also prove it using Tschirnhaus Transformation.



                                                      Let $,x=y-frac{b}{2a}$.$,$ Then $,ax^2+bx+c=0,$ becomes
                                                      $$begin{align}ay^2-frac{b^2}{4a}+c&=0\
                                                      iff y^2&=frac{b^2-4ac}{4a^2}\
                                                      iff y&=frac{pmsqrt{b^2-4ac}}{2a}\
                                                      iff x&=frac{pmsqrt{b^2-4ac}}{2a}-frac{b}{2a}\
                                                      iff x&=frac{-bpmsqrt{b^2-4ac}}{2a}end{align}$$





                                                      In general, the substitution for $, a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0,$ is $,x=y-frac{a_{n-1}}{na_n}$,



                                                      which leaves the polynomial without a term of degree $n-1$.







                                                      share|cite|improve this answer














                                                      share|cite|improve this answer



                                                      share|cite|improve this answer








                                                      edited May 18 '15 at 21:30

























                                                      answered Feb 3 '15 at 18:32









                                                      user26486user26486

                                                      9,29021948




                                                      9,29021948












                                                      • $begingroup$
                                                        Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                        $endgroup$
                                                        – Daniel W. Farlow
                                                        Feb 8 '15 at 4:19






                                                      • 1




                                                        $begingroup$
                                                        @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                        $endgroup$
                                                        – user26486
                                                        Feb 8 '15 at 4:48








                                                      • 3




                                                        $begingroup$
                                                        I agree with @induktio
                                                        $endgroup$
                                                        – fancynancy
                                                        Feb 8 '15 at 5:28






                                                      • 3




                                                        $begingroup$
                                                        I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                        $endgroup$
                                                        – user26486
                                                        Feb 19 '15 at 15:01


















                                                      • $begingroup$
                                                        Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                        $endgroup$
                                                        – Daniel W. Farlow
                                                        Feb 8 '15 at 4:19






                                                      • 1




                                                        $begingroup$
                                                        @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                        $endgroup$
                                                        – user26486
                                                        Feb 8 '15 at 4:48








                                                      • 3




                                                        $begingroup$
                                                        I agree with @induktio
                                                        $endgroup$
                                                        – fancynancy
                                                        Feb 8 '15 at 5:28






                                                      • 3




                                                        $begingroup$
                                                        I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                        $endgroup$
                                                        – user26486
                                                        Feb 19 '15 at 15:01
















                                                      $begingroup$
                                                      Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                      $endgroup$
                                                      – Daniel W. Farlow
                                                      Feb 8 '15 at 4:19




                                                      $begingroup$
                                                      Hardly adequate. Clearly the OP was looking for more than a standard derivation.
                                                      $endgroup$
                                                      – Daniel W. Farlow
                                                      Feb 8 '15 at 4:19




                                                      1




                                                      1




                                                      $begingroup$
                                                      @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                      $endgroup$
                                                      – user26486
                                                      Feb 8 '15 at 4:48






                                                      $begingroup$
                                                      @induktio What do you mean? The OP asked for a proof (how and why it is true) and do you see other answers not giving just proofs (e.g., Andre Nicolas answer, which merely gave a proof with a little justification why it is a better proof than the commonly taught one)?
                                                      $endgroup$
                                                      – user26486
                                                      Feb 8 '15 at 4:48






                                                      3




                                                      3




                                                      $begingroup$
                                                      I agree with @induktio
                                                      $endgroup$
                                                      – fancynancy
                                                      Feb 8 '15 at 5:28




                                                      $begingroup$
                                                      I agree with @induktio
                                                      $endgroup$
                                                      – fancynancy
                                                      Feb 8 '15 at 5:28




                                                      3




                                                      3




                                                      $begingroup$
                                                      I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                      $endgroup$
                                                      – user26486
                                                      Feb 19 '15 at 15:01




                                                      $begingroup$
                                                      I am very confused about the downvotes. OP asked for a proof, and I gave a proof that no one else has given here yet. Andre Nicolas too simply proved it just by multiplying both sides by $4a$ and completing the square, and yet he got all the upvotes while I got downvoted to oblivion.
                                                      $endgroup$
                                                      – user26486
                                                      Feb 19 '15 at 15:01











                                                      7












                                                      $begingroup$

                                                      Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.



                                                      Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $alpha$ and $beta$.
                                                      Now, $$~~~~~~~~~~~~~~~~~~~~~x-alpha = x-beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$
                                                      $$~~~~~~~~~~~~~~~~~~~~~k(x-alpha)(x-beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
                                                      Equating equation 1 and 2 (k is a constant),
                                                      $$ax^{2}+b{x}+c=k(x-alpha)(x-beta)$$
                                                      $$ax^{2}+b{x}+c=k(x^{2}-alpha x-beta x+alpha beta)$$
                                                      $$ax^{2}+b{x}+c=kx^{2}-k(alpha+beta )x+kalpha beta)$$
                                                      Comparing both sides, we get $$a=k~;~b=-k(alpha +beta)~;~c=kalpha beta$$
                                                      From this, we get $$alpha + beta = frac{-b}{a}~~;~~~alpha beta = frac{c}{a}$$



                                                      Now, to get the value of $alpha$, we follow the following procedure :
                                                      First we take out the value of $alpha - beta$, so that we can eliminate one term and find out the value of another.
                                                      $$(alpha-beta)^{2} = alpha ^{2}+ beta ^{2} - 2 alpha beta$$
                                                      Now we'll add $4 alpha beta $ on both the sides
                                                      $$(alpha-beta)^{2} +4 alpha beta = alpha ^{2}+ beta ^{2} + 2 alpha beta$$
                                                      $$(alpha-beta)^{2} +4 alpha beta = (alpha + beta )^{2} $$
                                                      $$(alpha-beta)^{2} = (alpha + beta )^{2} -4 alpha beta $$
                                                      $$alpha-beta = pm sqrt{(alpha + beta )^{2} -4 alpha beta } $$
                                                      Substituting the values of $alpha + beta$ and $alpha beta$, we get,
                                                      $$alpha-beta = pm sqrt{(frac{-b}{a} )^{2} -frac{4c}{a} } $$
                                                      $$alpha-beta = pm sqrt{frac{b^{2}-4ac}{a^{2}} } $$ or
                                                      $$~~~~~~~~~~~~~~~~~~~~~alpha-beta = frac{pm sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
                                                      Adding $Eq^{n} (2)~and~(3)$, we get,
                                                      $$2 alpha = frac{-b pm sqrt{b^{2}-4ac}}{a}$$
                                                      $$alpha = frac{-b pm sqrt{b^{2}-4ac}}{2a}$$






                                                      share|cite|improve this answer









                                                      $endgroup$









                                                      • 2




                                                        $begingroup$
                                                        $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                        $endgroup$
                                                        – miracle173
                                                        Nov 25 '14 at 5:16












                                                      • $begingroup$
                                                        you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                        $endgroup$
                                                        – user173094
                                                        Nov 25 '14 at 11:35
















                                                      7












                                                      $begingroup$

                                                      Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.



                                                      Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $alpha$ and $beta$.
                                                      Now, $$~~~~~~~~~~~~~~~~~~~~~x-alpha = x-beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$
                                                      $$~~~~~~~~~~~~~~~~~~~~~k(x-alpha)(x-beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
                                                      Equating equation 1 and 2 (k is a constant),
                                                      $$ax^{2}+b{x}+c=k(x-alpha)(x-beta)$$
                                                      $$ax^{2}+b{x}+c=k(x^{2}-alpha x-beta x+alpha beta)$$
                                                      $$ax^{2}+b{x}+c=kx^{2}-k(alpha+beta )x+kalpha beta)$$
                                                      Comparing both sides, we get $$a=k~;~b=-k(alpha +beta)~;~c=kalpha beta$$
                                                      From this, we get $$alpha + beta = frac{-b}{a}~~;~~~alpha beta = frac{c}{a}$$



                                                      Now, to get the value of $alpha$, we follow the following procedure :
                                                      First we take out the value of $alpha - beta$, so that we can eliminate one term and find out the value of another.
                                                      $$(alpha-beta)^{2} = alpha ^{2}+ beta ^{2} - 2 alpha beta$$
                                                      Now we'll add $4 alpha beta $ on both the sides
                                                      $$(alpha-beta)^{2} +4 alpha beta = alpha ^{2}+ beta ^{2} + 2 alpha beta$$
                                                      $$(alpha-beta)^{2} +4 alpha beta = (alpha + beta )^{2} $$
                                                      $$(alpha-beta)^{2} = (alpha + beta )^{2} -4 alpha beta $$
                                                      $$alpha-beta = pm sqrt{(alpha + beta )^{2} -4 alpha beta } $$
                                                      Substituting the values of $alpha + beta$ and $alpha beta$, we get,
                                                      $$alpha-beta = pm sqrt{(frac{-b}{a} )^{2} -frac{4c}{a} } $$
                                                      $$alpha-beta = pm sqrt{frac{b^{2}-4ac}{a^{2}} } $$ or
                                                      $$~~~~~~~~~~~~~~~~~~~~~alpha-beta = frac{pm sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
                                                      Adding $Eq^{n} (2)~and~(3)$, we get,
                                                      $$2 alpha = frac{-b pm sqrt{b^{2}-4ac}}{a}$$
                                                      $$alpha = frac{-b pm sqrt{b^{2}-4ac}}{2a}$$






                                                      share|cite|improve this answer









                                                      $endgroup$









                                                      • 2




                                                        $begingroup$
                                                        $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                        $endgroup$
                                                        – miracle173
                                                        Nov 25 '14 at 5:16












                                                      • $begingroup$
                                                        you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                        $endgroup$
                                                        – user173094
                                                        Nov 25 '14 at 11:35














                                                      7












                                                      7








                                                      7





                                                      $begingroup$

                                                      Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.



                                                      Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $alpha$ and $beta$.
                                                      Now, $$~~~~~~~~~~~~~~~~~~~~~x-alpha = x-beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$
                                                      $$~~~~~~~~~~~~~~~~~~~~~k(x-alpha)(x-beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
                                                      Equating equation 1 and 2 (k is a constant),
                                                      $$ax^{2}+b{x}+c=k(x-alpha)(x-beta)$$
                                                      $$ax^{2}+b{x}+c=k(x^{2}-alpha x-beta x+alpha beta)$$
                                                      $$ax^{2}+b{x}+c=kx^{2}-k(alpha+beta )x+kalpha beta)$$
                                                      Comparing both sides, we get $$a=k~;~b=-k(alpha +beta)~;~c=kalpha beta$$
                                                      From this, we get $$alpha + beta = frac{-b}{a}~~;~~~alpha beta = frac{c}{a}$$



                                                      Now, to get the value of $alpha$, we follow the following procedure :
                                                      First we take out the value of $alpha - beta$, so that we can eliminate one term and find out the value of another.
                                                      $$(alpha-beta)^{2} = alpha ^{2}+ beta ^{2} - 2 alpha beta$$
                                                      Now we'll add $4 alpha beta $ on both the sides
                                                      $$(alpha-beta)^{2} +4 alpha beta = alpha ^{2}+ beta ^{2} + 2 alpha beta$$
                                                      $$(alpha-beta)^{2} +4 alpha beta = (alpha + beta )^{2} $$
                                                      $$(alpha-beta)^{2} = (alpha + beta )^{2} -4 alpha beta $$
                                                      $$alpha-beta = pm sqrt{(alpha + beta )^{2} -4 alpha beta } $$
                                                      Substituting the values of $alpha + beta$ and $alpha beta$, we get,
                                                      $$alpha-beta = pm sqrt{(frac{-b}{a} )^{2} -frac{4c}{a} } $$
                                                      $$alpha-beta = pm sqrt{frac{b^{2}-4ac}{a^{2}} } $$ or
                                                      $$~~~~~~~~~~~~~~~~~~~~~alpha-beta = frac{pm sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
                                                      Adding $Eq^{n} (2)~and~(3)$, we get,
                                                      $$2 alpha = frac{-b pm sqrt{b^{2}-4ac}}{a}$$
                                                      $$alpha = frac{-b pm sqrt{b^{2}-4ac}}{2a}$$






                                                      share|cite|improve this answer









                                                      $endgroup$



                                                      Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.



                                                      Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $alpha$ and $beta$.
                                                      Now, $$~~~~~~~~~~~~~~~~~~~~~x-alpha = x-beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$
                                                      $$~~~~~~~~~~~~~~~~~~~~~k(x-alpha)(x-beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$
                                                      Equating equation 1 and 2 (k is a constant),
                                                      $$ax^{2}+b{x}+c=k(x-alpha)(x-beta)$$
                                                      $$ax^{2}+b{x}+c=k(x^{2}-alpha x-beta x+alpha beta)$$
                                                      $$ax^{2}+b{x}+c=kx^{2}-k(alpha+beta )x+kalpha beta)$$
                                                      Comparing both sides, we get $$a=k~;~b=-k(alpha +beta)~;~c=kalpha beta$$
                                                      From this, we get $$alpha + beta = frac{-b}{a}~~;~~~alpha beta = frac{c}{a}$$



                                                      Now, to get the value of $alpha$, we follow the following procedure :
                                                      First we take out the value of $alpha - beta$, so that we can eliminate one term and find out the value of another.
                                                      $$(alpha-beta)^{2} = alpha ^{2}+ beta ^{2} - 2 alpha beta$$
                                                      Now we'll add $4 alpha beta $ on both the sides
                                                      $$(alpha-beta)^{2} +4 alpha beta = alpha ^{2}+ beta ^{2} + 2 alpha beta$$
                                                      $$(alpha-beta)^{2} +4 alpha beta = (alpha + beta )^{2} $$
                                                      $$(alpha-beta)^{2} = (alpha + beta )^{2} -4 alpha beta $$
                                                      $$alpha-beta = pm sqrt{(alpha + beta )^{2} -4 alpha beta } $$
                                                      Substituting the values of $alpha + beta$ and $alpha beta$, we get,
                                                      $$alpha-beta = pm sqrt{(frac{-b}{a} )^{2} -frac{4c}{a} } $$
                                                      $$alpha-beta = pm sqrt{frac{b^{2}-4ac}{a^{2}} } $$ or
                                                      $$~~~~~~~~~~~~~~~~~~~~~alpha-beta = frac{pm sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$
                                                      Adding $Eq^{n} (2)~and~(3)$, we get,
                                                      $$2 alpha = frac{-b pm sqrt{b^{2}-4ac}}{a}$$
                                                      $$alpha = frac{-b pm sqrt{b^{2}-4ac}}{2a}$$







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered Oct 25 '14 at 7:35







                                                      user173094















                                                      • 2




                                                        $begingroup$
                                                        $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                        $endgroup$
                                                        – miracle173
                                                        Nov 25 '14 at 5:16












                                                      • $begingroup$
                                                        you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                        $endgroup$
                                                        – user173094
                                                        Nov 25 '14 at 11:35














                                                      • 2




                                                        $begingroup$
                                                        $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                        $endgroup$
                                                        – miracle173
                                                        Nov 25 '14 at 5:16












                                                      • $begingroup$
                                                        you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                        $endgroup$
                                                        – user173094
                                                        Nov 25 '14 at 11:35








                                                      2




                                                      2




                                                      $begingroup$
                                                      $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                      $endgroup$
                                                      – miracle173
                                                      Nov 25 '14 at 5:16






                                                      $begingroup$
                                                      $x-alpha = x-beta = 0$ would imply $alpha=beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring.
                                                      $endgroup$
                                                      – miracle173
                                                      Nov 25 '14 at 5:16














                                                      $begingroup$
                                                      you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                      $endgroup$
                                                      – user173094
                                                      Nov 25 '14 at 11:35




                                                      $begingroup$
                                                      you could rather write it as $x_1-alpha = x_2-beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated.
                                                      $endgroup$
                                                      – user173094
                                                      Nov 25 '14 at 11:35











                                                      7












                                                      $begingroup$

                                                      I'll just let the algebra speak for itself.



                                                      For $ a ne 0 $



                                                      If



                                                      $ax^2 = bx + c $



                                                      Then



                                                      $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                      PROOF:



                                                      $ 4aax^2 = 4abx + 4ac $



                                                      $(2ax)^2 + b^2 = 4abx + 4ac + b^2 $



                                                      $(2ax)^2 - 4abx + b^2 = b^2 + 4ac $



                                                      $ (2ax - b)^2 = b^2 + 4ac $



                                                      $ 2ax - b = pm sqrt{b^2 + 4ac} $



                                                      $2ax = b pm sqrt{b^2 + 4ac} $



                                                      $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                      Q.E.D.



                                                      SHORTCUT



                                                      Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.



                                                      For $ a ne 0 $



                                                      If



                                                      $ ax^2 + bx + c = 0 $



                                                      Then



                                                      $ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} $



                                                      Make the replacements in the 'if'



                                                      $ ax^2 + (-b)x + (-c) = 0 $



                                                      $ ax^2 = bx + c $



                                                      Make the replacements in the 'then'



                                                      $x = frac{-(-b) pm sqrt{(-b)^2 - 4a(-c)}}{2a} $



                                                      $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                      Q.E.D.






                                                      share|cite|improve this answer











                                                      $endgroup$


















                                                        7












                                                        $begingroup$

                                                        I'll just let the algebra speak for itself.



                                                        For $ a ne 0 $



                                                        If



                                                        $ax^2 = bx + c $



                                                        Then



                                                        $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                        PROOF:



                                                        $ 4aax^2 = 4abx + 4ac $



                                                        $(2ax)^2 + b^2 = 4abx + 4ac + b^2 $



                                                        $(2ax)^2 - 4abx + b^2 = b^2 + 4ac $



                                                        $ (2ax - b)^2 = b^2 + 4ac $



                                                        $ 2ax - b = pm sqrt{b^2 + 4ac} $



                                                        $2ax = b pm sqrt{b^2 + 4ac} $



                                                        $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                        Q.E.D.



                                                        SHORTCUT



                                                        Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.



                                                        For $ a ne 0 $



                                                        If



                                                        $ ax^2 + bx + c = 0 $



                                                        Then



                                                        $ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} $



                                                        Make the replacements in the 'if'



                                                        $ ax^2 + (-b)x + (-c) = 0 $



                                                        $ ax^2 = bx + c $



                                                        Make the replacements in the 'then'



                                                        $x = frac{-(-b) pm sqrt{(-b)^2 - 4a(-c)}}{2a} $



                                                        $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                        Q.E.D.






                                                        share|cite|improve this answer











                                                        $endgroup$
















                                                          7












                                                          7








                                                          7





                                                          $begingroup$

                                                          I'll just let the algebra speak for itself.



                                                          For $ a ne 0 $



                                                          If



                                                          $ax^2 = bx + c $



                                                          Then



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          PROOF:



                                                          $ 4aax^2 = 4abx + 4ac $



                                                          $(2ax)^2 + b^2 = 4abx + 4ac + b^2 $



                                                          $(2ax)^2 - 4abx + b^2 = b^2 + 4ac $



                                                          $ (2ax - b)^2 = b^2 + 4ac $



                                                          $ 2ax - b = pm sqrt{b^2 + 4ac} $



                                                          $2ax = b pm sqrt{b^2 + 4ac} $



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          Q.E.D.



                                                          SHORTCUT



                                                          Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.



                                                          For $ a ne 0 $



                                                          If



                                                          $ ax^2 + bx + c = 0 $



                                                          Then



                                                          $ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} $



                                                          Make the replacements in the 'if'



                                                          $ ax^2 + (-b)x + (-c) = 0 $



                                                          $ ax^2 = bx + c $



                                                          Make the replacements in the 'then'



                                                          $x = frac{-(-b) pm sqrt{(-b)^2 - 4a(-c)}}{2a} $



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          Q.E.D.






                                                          share|cite|improve this answer











                                                          $endgroup$



                                                          I'll just let the algebra speak for itself.



                                                          For $ a ne 0 $



                                                          If



                                                          $ax^2 = bx + c $



                                                          Then



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          PROOF:



                                                          $ 4aax^2 = 4abx + 4ac $



                                                          $(2ax)^2 + b^2 = 4abx + 4ac + b^2 $



                                                          $(2ax)^2 - 4abx + b^2 = b^2 + 4ac $



                                                          $ (2ax - b)^2 = b^2 + 4ac $



                                                          $ 2ax - b = pm sqrt{b^2 + 4ac} $



                                                          $2ax = b pm sqrt{b^2 + 4ac} $



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          Q.E.D.



                                                          SHORTCUT



                                                          Using the standard full definition , replace b with -b and c with -c to get the formula I derived in this post.



                                                          For $ a ne 0 $



                                                          If



                                                          $ ax^2 + bx + c = 0 $



                                                          Then



                                                          $ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} $



                                                          Make the replacements in the 'if'



                                                          $ ax^2 + (-b)x + (-c) = 0 $



                                                          $ ax^2 = bx + c $



                                                          Make the replacements in the 'then'



                                                          $x = frac{-(-b) pm sqrt{(-b)^2 - 4a(-c)}}{2a} $



                                                          $ x = frac{b pm sqrt{b^2 + 4ac}}{2a} $



                                                          Q.E.D.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited Jul 20 '16 at 11:38

























                                                          answered Jan 30 '14 at 19:29









                                                          neofoxmulderneofoxmulder

                                                          984616




                                                          984616























                                                              5












                                                              $begingroup$

                                                              The proof of the quadratic formula takes advantage of completing the square.
                                                              $$ax^2 + bx + c = 0, a neq 0, a, b, c in mathbb R$$
                                                              $$ax^2 + bx = -c$$
                                                              $$x^2 + dfrac{b}{a}x = -dfrac{c}{a}$$
                                                              $$x^2 + dfrac{b}{a}x + left(dfrac{b}{2a}right)^2 = -dfrac{c}{a} + left(dfrac{b}{2a}right)^2$$
                                                              $$left(x+dfrac{b}{2a}right)^2 = -dfrac{c}{a} + dfrac{b^2}{4a^2}$$
                                                              $$left(x+dfrac{b}{2a}right)^2 = -dfrac{4ac}{4a^2} + dfrac{b^2}{4a^2}$$
                                                              $$left(x+dfrac{b}{2a}right)^2 = dfrac{-4ac+b^2}{4a^2}$$
                                                              $$left(x+dfrac{b}{2a}right)^2 = dfrac{b^2-4ac}{4a^2}$$
                                                              $$x+dfrac{b}{2a} = pm sqrt{dfrac{b^2-4ac}{4a^2}}$$
                                                              $$x+dfrac{b}{2a} = pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                              $$x=-dfrac{b}{2a} pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                              $$boxed{x=dfrac{-b pm sqrt{b^2-4ac}}{2a}}$$
                                                              This concludes the proof.






                                                              share|cite|improve this answer









                                                              $endgroup$


















                                                                5












                                                                $begingroup$

                                                                The proof of the quadratic formula takes advantage of completing the square.
                                                                $$ax^2 + bx + c = 0, a neq 0, a, b, c in mathbb R$$
                                                                $$ax^2 + bx = -c$$
                                                                $$x^2 + dfrac{b}{a}x = -dfrac{c}{a}$$
                                                                $$x^2 + dfrac{b}{a}x + left(dfrac{b}{2a}right)^2 = -dfrac{c}{a} + left(dfrac{b}{2a}right)^2$$
                                                                $$left(x+dfrac{b}{2a}right)^2 = -dfrac{c}{a} + dfrac{b^2}{4a^2}$$
                                                                $$left(x+dfrac{b}{2a}right)^2 = -dfrac{4ac}{4a^2} + dfrac{b^2}{4a^2}$$
                                                                $$left(x+dfrac{b}{2a}right)^2 = dfrac{-4ac+b^2}{4a^2}$$
                                                                $$left(x+dfrac{b}{2a}right)^2 = dfrac{b^2-4ac}{4a^2}$$
                                                                $$x+dfrac{b}{2a} = pm sqrt{dfrac{b^2-4ac}{4a^2}}$$
                                                                $$x+dfrac{b}{2a} = pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                $$x=-dfrac{b}{2a} pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                $$boxed{x=dfrac{-b pm sqrt{b^2-4ac}}{2a}}$$
                                                                This concludes the proof.






                                                                share|cite|improve this answer









                                                                $endgroup$
















                                                                  5












                                                                  5








                                                                  5





                                                                  $begingroup$

                                                                  The proof of the quadratic formula takes advantage of completing the square.
                                                                  $$ax^2 + bx + c = 0, a neq 0, a, b, c in mathbb R$$
                                                                  $$ax^2 + bx = -c$$
                                                                  $$x^2 + dfrac{b}{a}x = -dfrac{c}{a}$$
                                                                  $$x^2 + dfrac{b}{a}x + left(dfrac{b}{2a}right)^2 = -dfrac{c}{a} + left(dfrac{b}{2a}right)^2$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = -dfrac{c}{a} + dfrac{b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = -dfrac{4ac}{4a^2} + dfrac{b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = dfrac{-4ac+b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = dfrac{b^2-4ac}{4a^2}$$
                                                                  $$x+dfrac{b}{2a} = pm sqrt{dfrac{b^2-4ac}{4a^2}}$$
                                                                  $$x+dfrac{b}{2a} = pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                  $$x=-dfrac{b}{2a} pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                  $$boxed{x=dfrac{-b pm sqrt{b^2-4ac}}{2a}}$$
                                                                  This concludes the proof.






                                                                  share|cite|improve this answer









                                                                  $endgroup$



                                                                  The proof of the quadratic formula takes advantage of completing the square.
                                                                  $$ax^2 + bx + c = 0, a neq 0, a, b, c in mathbb R$$
                                                                  $$ax^2 + bx = -c$$
                                                                  $$x^2 + dfrac{b}{a}x = -dfrac{c}{a}$$
                                                                  $$x^2 + dfrac{b}{a}x + left(dfrac{b}{2a}right)^2 = -dfrac{c}{a} + left(dfrac{b}{2a}right)^2$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = -dfrac{c}{a} + dfrac{b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = -dfrac{4ac}{4a^2} + dfrac{b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = dfrac{-4ac+b^2}{4a^2}$$
                                                                  $$left(x+dfrac{b}{2a}right)^2 = dfrac{b^2-4ac}{4a^2}$$
                                                                  $$x+dfrac{b}{2a} = pm sqrt{dfrac{b^2-4ac}{4a^2}}$$
                                                                  $$x+dfrac{b}{2a} = pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                  $$x=-dfrac{b}{2a} pm dfrac{sqrt{b^2-4ac}}{2a}$$
                                                                  $$boxed{x=dfrac{-b pm sqrt{b^2-4ac}}{2a}}$$
                                                                  This concludes the proof.







                                                                  share|cite|improve this answer












                                                                  share|cite|improve this answer



                                                                  share|cite|improve this answer










                                                                  answered Feb 21 '14 at 7:10









                                                                  TrueDefaultTrueDefault

                                                                  3,77672849




                                                                  3,77672849























                                                                      5












                                                                      $begingroup$

                                                                      The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=pm sqrt{a}$.



                                                                      The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+frac{b}{a} x + frac{c}{a}=0$. Writing $p=frac{b}{a}$, $q=frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-frac{p}{2}$, then a little calculation shows $y^2 - frac{p^2}{4}+q=0$.



                                                                      $$Rightarrow ~ y=pm sqrt{frac{p^2}{4}-q} ~ Rightarrow ~ x = -frac{p}{2} pm sqrt{frac{p^2}{4}-q} = frac{-b pm sqrt{b^2-4ac}}{2a}.$$



                                                                      Geometrically, $x=y-frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.



                                                                      enter image description here



                                                                      This method also works for polynomials of higher degrees: Tschirnhaus transformation.






                                                                      share|cite|improve this answer











                                                                      $endgroup$













                                                                      • $begingroup$
                                                                        it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                        $endgroup$
                                                                        – user153330
                                                                        Mar 21 '16 at 20:04


















                                                                      5












                                                                      $begingroup$

                                                                      The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=pm sqrt{a}$.



                                                                      The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+frac{b}{a} x + frac{c}{a}=0$. Writing $p=frac{b}{a}$, $q=frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-frac{p}{2}$, then a little calculation shows $y^2 - frac{p^2}{4}+q=0$.



                                                                      $$Rightarrow ~ y=pm sqrt{frac{p^2}{4}-q} ~ Rightarrow ~ x = -frac{p}{2} pm sqrt{frac{p^2}{4}-q} = frac{-b pm sqrt{b^2-4ac}}{2a}.$$



                                                                      Geometrically, $x=y-frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.



                                                                      enter image description here



                                                                      This method also works for polynomials of higher degrees: Tschirnhaus transformation.






                                                                      share|cite|improve this answer











                                                                      $endgroup$













                                                                      • $begingroup$
                                                                        it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                        $endgroup$
                                                                        – user153330
                                                                        Mar 21 '16 at 20:04
















                                                                      5












                                                                      5








                                                                      5





                                                                      $begingroup$

                                                                      The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=pm sqrt{a}$.



                                                                      The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+frac{b}{a} x + frac{c}{a}=0$. Writing $p=frac{b}{a}$, $q=frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-frac{p}{2}$, then a little calculation shows $y^2 - frac{p^2}{4}+q=0$.



                                                                      $$Rightarrow ~ y=pm sqrt{frac{p^2}{4}-q} ~ Rightarrow ~ x = -frac{p}{2} pm sqrt{frac{p^2}{4}-q} = frac{-b pm sqrt{b^2-4ac}}{2a}.$$



                                                                      Geometrically, $x=y-frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.



                                                                      enter image description here



                                                                      This method also works for polynomials of higher degrees: Tschirnhaus transformation.






                                                                      share|cite|improve this answer











                                                                      $endgroup$



                                                                      The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=pm sqrt{a}$.



                                                                      The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+frac{b}{a} x + frac{c}{a}=0$. Writing $p=frac{b}{a}$, $q=frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-frac{p}{2}$, then a little calculation shows $y^2 - frac{p^2}{4}+q=0$.



                                                                      $$Rightarrow ~ y=pm sqrt{frac{p^2}{4}-q} ~ Rightarrow ~ x = -frac{p}{2} pm sqrt{frac{p^2}{4}-q} = frac{-b pm sqrt{b^2-4ac}}{2a}.$$



                                                                      Geometrically, $x=y-frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.



                                                                      enter image description here



                                                                      This method also works for polynomials of higher degrees: Tschirnhaus transformation.







                                                                      share|cite|improve this answer














                                                                      share|cite|improve this answer



                                                                      share|cite|improve this answer








                                                                      edited Nov 14 '14 at 22:46

























                                                                      answered Jul 26 '14 at 9:25









                                                                      Martin BrandenburgMartin Brandenburg

                                                                      108k13158329




                                                                      108k13158329












                                                                      • $begingroup$
                                                                        it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                        $endgroup$
                                                                        – user153330
                                                                        Mar 21 '16 at 20:04




















                                                                      • $begingroup$
                                                                        it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                        $endgroup$
                                                                        – user153330
                                                                        Mar 21 '16 at 20:04


















                                                                      $begingroup$
                                                                      it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                      $endgroup$
                                                                      – user153330
                                                                      Mar 21 '16 at 20:04






                                                                      $begingroup$
                                                                      it would be awesome if you give a way to see how that transformation arose (or how one can come up with it in the first place)
                                                                      $endgroup$
                                                                      – user153330
                                                                      Mar 21 '16 at 20:04













                                                                      4












                                                                      $begingroup$

                                                                      I've seen several "proofs" of this formula, all of which proceed by a sequence of randomly-chosen algebraic operations and end at the famous formula. Of course, no normal human being would ever pluck these out of thin air like that.



                                                                      For me, the realization was this: Consider the formula



                                                                      $$(x - alpha)(x - beta) = 0$$



                                                                      If $x=alpha$, then $x-alpha$ is obviously zero. Whatever $x-beta$ happens to be, it gets multiplied by zero, so the result is still zero. So in other words, one solution to this equation is $x=alpha$. By a similar line of reasoning, $x=beta$ is also a solution. So here we have purposely constructed an equation that has exactly two solutions: $alpha$ and $beta$ (which can be whatever we choose them to be).



                                                                      Now let's open the brackets:



                                                                      $$(x-alpha)(x-beta) = 0$$
                                                                      $$x(x-beta) - alpha(x-beta) = 0$$
                                                                      $$x^2 - beta x - alpha x + alpha beta = 0$$
                                                                      $$x^2 - (alpha + beta)x + (alpha beta) = 0$$



                                                                      Now suppose we have an equation that looks like



                                                                      $$x^2 + Ax + B = 0$$



                                                                      (Notice that this isn't the usual general quadratic equation. I've assumed that the $x^2$ has a coefficient of 1.)



                                                                      If we line the two equations up, it becomes clear that



                                                                      $$A = -(alpha + beta)$$
                                                                      $$B = (alpha beta)$$



                                                                      This is the direct relationship between the coefficients we can see, and the solutions we want to find. Now, how to figure out the solutions from the coefficients?



                                                                      We know that $-A$ is the sum of the solutions, and $B$ is the product of the solutions. But what are the solutions themselves? Hmm.



                                                                      Well, if $-A$ is the sum, then $-A/2$ is the arithmetic mean of the solutions. I.e., it's exactly half way between the two solutions. Great, if we can just figure out how far apart the solutions are, we could exactly compute them!



                                                                      So we have the sum and the product, but we want the difference. Hmm. OK. But how do we do that?



                                                                      We've got $alpha + beta$ and we've got $alpha beta$, but we want $alpha - beta$. Tricky…



                                                                      The solution is Black Magic. Observe: The thing we have is $alpha + beta$, and



                                                                      $$(alpha + beta)^2 = alpha^2 + 2 alpha beta + beta^2$$



                                                                      The thing we want is $alpha - beta$, and



                                                                      $$(alpha - beta)^2 = alpha^2 - 2 alpha beta + beta^2$$



                                                                      The difference between these two expressions is exactly $4alphabeta$which we can compute! If $B = alphabeta$, then $4B = 4alphabeta$. What I am saying is that



                                                                      $$(alpha - beta)^2 = (alpha + beta)^2 - 4alphabeta = A^2 - 4B$$



                                                                      It therefore follows that



                                                                      $$alpha - beta = sqrt{A^2 - 4B}$$



                                                                      Recalling that $-A$ is the sum of the solutions and so $-A/2$ is the average, we have



                                                                      $$alpha = -frac{A}{2} - frac{sqrt{A^2 - 4B}}{2}$$
                                                                      $$beta = -frac{A}{2} + frac{sqrt{A^2 - 4B}}{2}$$



                                                                      In other words, half the sum minus half the difference gives the lower solution, and half the sum plus half the difference gives the upper solution.



                                                                      $$x = frac{-A pm sqrt{A^2 - 4B}}{2}$$



                                                                      Notice that this doesn't look quite like the familiar formula, because we assumed that the $x^2$ coefficient equals exactly 1. But notice that if we divide all the coefficients by the leading one, we can get this form!



                                                                      $$ax^2 + bx + c = 0$$
                                                                      $$x^2 + frac{b}{a}x + frac{c}{a} = 0$$
                                                                      $$A = frac{b}{a}$$
                                                                      $$B = frac{c}{a}$$



                                                                      Substitute these into our previous formula, and (trust me) the familiar quadratic solution formula pops out.





                                                                      Notice that we can do this trick for a polynomial of any degree!



                                                                      $$(x - alpha)(x - beta)(x - gamma) = 0$$



                                                                      This gives a cubic equation with known solutions. Opening the brackets,



                                                                      $$x^3 - (alpha + beta + gamma)x^2 + (alphabeta + alphagamma + betagamma)x - (alpha beta gamma) = 0$$



                                                                      Given the equation



                                                                      $$x^3 + Ax^2 + Bx + C = 0$$



                                                                      we get



                                                                      $$A = -(alpha + beta + gamma)$$
                                                                      $$B = (alphabeta + alphagamma + betagamma)$$
                                                                      $$C = -(alphabetagamma)$$



                                                                      The trouble is, now we need to somehow compute $alpha$, $beta$ and $gamma$ given only $A$, $B$ and $C$. And it turns out this is way harder than last time. (Indeed, I have no damn idea how to do it! Computing the average doesn't really help this time.)






                                                                      share|cite|improve this answer











                                                                      $endgroup$


















                                                                        4












                                                                        $begingroup$

                                                                        I've seen several "proofs" of this formula, all of which proceed by a sequence of randomly-chosen algebraic operations and end at the famous formula. Of course, no normal human being would ever pluck these out of thin air like that.



                                                                        For me, the realization was this: Consider the formula



                                                                        $$(x - alpha)(x - beta) = 0$$



                                                                        If $x=alpha$, then $x-alpha$ is obviously zero. Whatever $x-beta$ happens to be, it gets multiplied by zero, so the result is still zero. So in other words, one solution to this equation is $x=alpha$. By a similar line of reasoning, $x=beta$ is also a solution. So here we have purposely constructed an equation that has exactly two solutions: $alpha$ and $beta$ (which can be whatever we choose them to be).



                                                                        Now let's open the brackets:



                                                                        $$(x-alpha)(x-beta) = 0$$
                                                                        $$x(x-beta) - alpha(x-beta) = 0$$
                                                                        $$x^2 - beta x - alpha x + alpha beta = 0$$
                                                                        $$x^2 - (alpha + beta)x + (alpha beta) = 0$$



                                                                        Now suppose we have an equation that looks like



                                                                        $$x^2 + Ax + B = 0$$



                                                                        (Notice that this isn't the usual general quadratic equation. I've assumed that the $x^2$ has a coefficient of 1.)



                                                                        If we line the two equations up, it becomes clear that



                                                                        $$A = -(alpha + beta)$$
                                                                        $$B = (alpha beta)$$



                                                                        This is the direct relationship between the coefficients we can see, and the solutions we want to find. Now, how to figure out the solutions from the coefficients?



                                                                        We know that $-A$ is the sum of the solutions, and $B$ is the product of the solutions. But what are the solutions themselves? Hmm.



                                                                        Well, if $-A$ is the sum, then $-A/2$ is the arithmetic mean of the solutions. I.e., it's exactly half way between the two solutions. Great, if we can just figure out how far apart the solutions are, we could exactly compute them!



                                                                        So we have the sum and the product, but we want the difference. Hmm. OK. But how do we do that?



                                                                        We've got $alpha + beta$ and we've got $alpha beta$, but we want $alpha - beta$. Tricky…



                                                                        The solution is Black Magic. Observe: The thing we have is $alpha + beta$, and



                                                                        $$(alpha + beta)^2 = alpha^2 + 2 alpha beta + beta^2$$



                                                                        The thing we want is $alpha - beta$, and



                                                                        $$(alpha - beta)^2 = alpha^2 - 2 alpha beta + beta^2$$



                                                                        The difference between these two expressions is exactly $4alphabeta$which we can compute! If $B = alphabeta$, then $4B = 4alphabeta$. What I am saying is that



                                                                        $$(alpha - beta)^2 = (alpha + beta)^2 - 4alphabeta = A^2 - 4B$$



                                                                        It therefore follows that



                                                                        $$alpha - beta = sqrt{A^2 - 4B}$$



                                                                        Recalling that $-A$ is the sum of the solutions and so $-A/2$ is the average, we have



                                                                        $$alpha = -frac{A}{2} - frac{sqrt{A^2 - 4B}}{2}$$
                                                                        $$beta = -frac{A}{2} + frac{sqrt{A^2 - 4B}}{2}$$



                                                                        In other words, half the sum minus half the difference gives the lower solution, and half the sum plus half the difference gives the upper solution.



                                                                        $$x = frac{-A pm sqrt{A^2 - 4B}}{2}$$



                                                                        Notice that this doesn't look quite like the familiar formula, because we assumed that the $x^2$ coefficient equals exactly 1. But notice that if we divide all the coefficients by the leading one, we can get this form!



                                                                        $$ax^2 + bx + c = 0$$
                                                                        $$x^2 + frac{b}{a}x + frac{c}{a} = 0$$
                                                                        $$A = frac{b}{a}$$
                                                                        $$B = frac{c}{a}$$



                                                                        Substitute these into our previous formula, and (trust me) the familiar quadratic solution formula pops out.





                                                                        Notice that we can do this trick for a polynomial of any degree!



                                                                        $$(x - alpha)(x - beta)(x - gamma) = 0$$



                                                                        This gives a cubic equation with known solutions. Opening the brackets,



                                                                        $$x^3 - (alpha + beta + gamma)x^2 + (alphabeta + alphagamma + betagamma)x - (alpha beta gamma) = 0$$



                                                                        Given the equation



                                                                        $$x^3 + Ax^2 + Bx + C = 0$$



                                                                        we get



                                                                        $$A = -(alpha + beta + gamma)$$
                                                                        $$B = (alphabeta + alphagamma + betagamma)$$
                                                                        $$C = -(alphabetagamma)$$



                                                                        The trouble is, now we need to somehow compute $alpha$, $beta$ and $gamma$ given only $A$, $B$ and $C$. And it turns out this is way harder than last time. (Indeed, I have no damn idea how to do it! Computing the average doesn't really help this time.)






                                                                        share|cite|improve this answer











                                                                        $endgroup$
















                                                                          4












                                                                          4








                                                                          4





                                                                          $begingroup$

                                                                          I've seen several "proofs" of this formula, all of which proceed by a sequence of randomly-chosen algebraic operations and end at the famous formula. Of course, no normal human being would ever pluck these out of thin air like that.



                                                                          For me, the realization was this: Consider the formula



                                                                          $$(x - alpha)(x - beta) = 0$$



                                                                          If $x=alpha$, then $x-alpha$ is obviously zero. Whatever $x-beta$ happens to be, it gets multiplied by zero, so the result is still zero. So in other words, one solution to this equation is $x=alpha$. By a similar line of reasoning, $x=beta$ is also a solution. So here we have purposely constructed an equation that has exactly two solutions: $alpha$ and $beta$ (which can be whatever we choose them to be).



                                                                          Now let's open the brackets:



                                                                          $$(x-alpha)(x-beta) = 0$$
                                                                          $$x(x-beta) - alpha(x-beta) = 0$$
                                                                          $$x^2 - beta x - alpha x + alpha beta = 0$$
                                                                          $$x^2 - (alpha + beta)x + (alpha beta) = 0$$



                                                                          Now suppose we have an equation that looks like



                                                                          $$x^2 + Ax + B = 0$$



                                                                          (Notice that this isn't the usual general quadratic equation. I've assumed that the $x^2$ has a coefficient of 1.)



                                                                          If we line the two equations up, it becomes clear that



                                                                          $$A = -(alpha + beta)$$
                                                                          $$B = (alpha beta)$$



                                                                          This is the direct relationship between the coefficients we can see, and the solutions we want to find. Now, how to figure out the solutions from the coefficients?



                                                                          We know that $-A$ is the sum of the solutions, and $B$ is the product of the solutions. But what are the solutions themselves? Hmm.



                                                                          Well, if $-A$ is the sum, then $-A/2$ is the arithmetic mean of the solutions. I.e., it's exactly half way between the two solutions. Great, if we can just figure out how far apart the solutions are, we could exactly compute them!



                                                                          So we have the sum and the product, but we want the difference. Hmm. OK. But how do we do that?



                                                                          We've got $alpha + beta$ and we've got $alpha beta$, but we want $alpha - beta$. Tricky…



                                                                          The solution is Black Magic. Observe: The thing we have is $alpha + beta$, and



                                                                          $$(alpha + beta)^2 = alpha^2 + 2 alpha beta + beta^2$$



                                                                          The thing we want is $alpha - beta$, and



                                                                          $$(alpha - beta)^2 = alpha^2 - 2 alpha beta + beta^2$$



                                                                          The difference between these two expressions is exactly $4alphabeta$which we can compute! If $B = alphabeta$, then $4B = 4alphabeta$. What I am saying is that



                                                                          $$(alpha - beta)^2 = (alpha + beta)^2 - 4alphabeta = A^2 - 4B$$



                                                                          It therefore follows that



                                                                          $$alpha - beta = sqrt{A^2 - 4B}$$



                                                                          Recalling that $-A$ is the sum of the solutions and so $-A/2$ is the average, we have



                                                                          $$alpha = -frac{A}{2} - frac{sqrt{A^2 - 4B}}{2}$$
                                                                          $$beta = -frac{A}{2} + frac{sqrt{A^2 - 4B}}{2}$$



                                                                          In other words, half the sum minus half the difference gives the lower solution, and half the sum plus half the difference gives the upper solution.



                                                                          $$x = frac{-A pm sqrt{A^2 - 4B}}{2}$$



                                                                          Notice that this doesn't look quite like the familiar formula, because we assumed that the $x^2$ coefficient equals exactly 1. But notice that if we divide all the coefficients by the leading one, we can get this form!



                                                                          $$ax^2 + bx + c = 0$$
                                                                          $$x^2 + frac{b}{a}x + frac{c}{a} = 0$$
                                                                          $$A = frac{b}{a}$$
                                                                          $$B = frac{c}{a}$$



                                                                          Substitute these into our previous formula, and (trust me) the familiar quadratic solution formula pops out.





                                                                          Notice that we can do this trick for a polynomial of any degree!



                                                                          $$(x - alpha)(x - beta)(x - gamma) = 0$$



                                                                          This gives a cubic equation with known solutions. Opening the brackets,



                                                                          $$x^3 - (alpha + beta + gamma)x^2 + (alphabeta + alphagamma + betagamma)x - (alpha beta gamma) = 0$$



                                                                          Given the equation



                                                                          $$x^3 + Ax^2 + Bx + C = 0$$



                                                                          we get



                                                                          $$A = -(alpha + beta + gamma)$$
                                                                          $$B = (alphabeta + alphagamma + betagamma)$$
                                                                          $$C = -(alphabetagamma)$$



                                                                          The trouble is, now we need to somehow compute $alpha$, $beta$ and $gamma$ given only $A$, $B$ and $C$. And it turns out this is way harder than last time. (Indeed, I have no damn idea how to do it! Computing the average doesn't really help this time.)






                                                                          share|cite|improve this answer











                                                                          $endgroup$



                                                                          I've seen several "proofs" of this formula, all of which proceed by a sequence of randomly-chosen algebraic operations and end at the famous formula. Of course, no normal human being would ever pluck these out of thin air like that.



                                                                          For me, the realization was this: Consider the formula



                                                                          $$(x - alpha)(x - beta) = 0$$



                                                                          If $x=alpha$, then $x-alpha$ is obviously zero. Whatever $x-beta$ happens to be, it gets multiplied by zero, so the result is still zero. So in other words, one solution to this equation is $x=alpha$. By a similar line of reasoning, $x=beta$ is also a solution. So here we have purposely constructed an equation that has exactly two solutions: $alpha$ and $beta$ (which can be whatever we choose them to be).



                                                                          Now let's open the brackets:



                                                                          $$(x-alpha)(x-beta) = 0$$
                                                                          $$x(x-beta) - alpha(x-beta) = 0$$
                                                                          $$x^2 - beta x - alpha x + alpha beta = 0$$
                                                                          $$x^2 - (alpha + beta)x + (alpha beta) = 0$$



                                                                          Now suppose we have an equation that looks like



                                                                          $$x^2 + Ax + B = 0$$



                                                                          (Notice that this isn't the usual general quadratic equation. I've assumed that the $x^2$ has a coefficient of 1.)



                                                                          If we line the two equations up, it becomes clear that



                                                                          $$A = -(alpha + beta)$$
                                                                          $$B = (alpha beta)$$



                                                                          This is the direct relationship between the coefficients we can see, and the solutions we want to find. Now, how to figure out the solutions from the coefficients?



                                                                          We know that $-A$ is the sum of the solutions, and $B$ is the product of the solutions. But what are the solutions themselves? Hmm.



                                                                          Well, if $-A$ is the sum, then $-A/2$ is the arithmetic mean of the solutions. I.e., it's exactly half way between the two solutions. Great, if we can just figure out how far apart the solutions are, we could exactly compute them!



                                                                          So we have the sum and the product, but we want the difference. Hmm. OK. But how do we do that?



                                                                          We've got $alpha + beta$ and we've got $alpha beta$, but we want $alpha - beta$. Tricky…



                                                                          The solution is Black Magic. Observe: The thing we have is $alpha + beta$, and



                                                                          $$(alpha + beta)^2 = alpha^2 + 2 alpha beta + beta^2$$



                                                                          The thing we want is $alpha - beta$, and



                                                                          $$(alpha - beta)^2 = alpha^2 - 2 alpha beta + beta^2$$



                                                                          The difference between these two expressions is exactly $4alphabeta$which we can compute! If $B = alphabeta$, then $4B = 4alphabeta$. What I am saying is that



                                                                          $$(alpha - beta)^2 = (alpha + beta)^2 - 4alphabeta = A^2 - 4B$$



                                                                          It therefore follows that



                                                                          $$alpha - beta = sqrt{A^2 - 4B}$$



                                                                          Recalling that $-A$ is the sum of the solutions and so $-A/2$ is the average, we have



                                                                          $$alpha = -frac{A}{2} - frac{sqrt{A^2 - 4B}}{2}$$
                                                                          $$beta = -frac{A}{2} + frac{sqrt{A^2 - 4B}}{2}$$



                                                                          In other words, half the sum minus half the difference gives the lower solution, and half the sum plus half the difference gives the upper solution.



                                                                          $$x = frac{-A pm sqrt{A^2 - 4B}}{2}$$



                                                                          Notice that this doesn't look quite like the familiar formula, because we assumed that the $x^2$ coefficient equals exactly 1. But notice that if we divide all the coefficients by the leading one, we can get this form!



                                                                          $$ax^2 + bx + c = 0$$
                                                                          $$x^2 + frac{b}{a}x + frac{c}{a} = 0$$
                                                                          $$A = frac{b}{a}$$
                                                                          $$B = frac{c}{a}$$



                                                                          Substitute these into our previous formula, and (trust me) the familiar quadratic solution formula pops out.





                                                                          Notice that we can do this trick for a polynomial of any degree!



                                                                          $$(x - alpha)(x - beta)(x - gamma) = 0$$



                                                                          This gives a cubic equation with known solutions. Opening the brackets,



                                                                          $$x^3 - (alpha + beta + gamma)x^2 + (alphabeta + alphagamma + betagamma)x - (alpha beta gamma) = 0$$



                                                                          Given the equation



                                                                          $$x^3 + Ax^2 + Bx + C = 0$$



                                                                          we get



                                                                          $$A = -(alpha + beta + gamma)$$
                                                                          $$B = (alphabeta + alphagamma + betagamma)$$
                                                                          $$C = -(alphabetagamma)$$



                                                                          The trouble is, now we need to somehow compute $alpha$, $beta$ and $gamma$ given only $A$, $B$ and $C$. And it turns out this is way harder than last time. (Indeed, I have no damn idea how to do it! Computing the average doesn't really help this time.)







                                                                          share|cite|improve this answer














                                                                          share|cite|improve this answer



                                                                          share|cite|improve this answer








                                                                          edited Dec 5 '18 at 9:13

























                                                                          answered Mar 10 '16 at 11:55









                                                                          MathematicalOrchidMathematicalOrchid

                                                                          3,01332034




                                                                          3,01332034























                                                                              3












                                                                              $begingroup$

                                                                              Graphical Approach
                                                                              The following method actually requires some knowledge about the graph of parabola and calculus. This is just another way to view the formula.
                                                                              For $a>0,,
                                                                              f^{prime}(x)=0,left[x=-frac{b}{2a}right]$ would give you the minima. Now substitute this in the expression to see whether it attains any negative value. If it is positive here then it wont have any real roots for obvious reasons given that you have sufficient knowledge about functions.
                                                                              Now if the function has real roots then the arithmetic mean of the roots would give the axis of parabola. The roots are symmetrical about this line so if we somehow find the difference between the roots and add half of it to $-frac{b}{2a}$ then we would get one root and subtracting half of difference would give the other root.We know the sum of roots is $-frac{b}{a}$ and product of roots is $frac{c}{a}$. So
                                                                              begin{equation}
                                                                              F+G=-frac{a}{b},,FG=frac{c}{a}$
                                                                              end{equation}
                                                                              therefore, $(F-G)^2 = left(-frac{a}{b}right)^2 -4frac{c}{a}$
                                                                              taking root,
                                                                              $|F-G|=left[left(-frac{a}{b}right)^2 -4frac{c}{a}right]^{frac{1}{2}}$
                                                                              therefore the roots are
                                                                              $-frac{b}{a} + frac{|F-G|}{2},-frac{b}{a} - frac{|F-G|}{2}$ enter image description here
                                                                              Substituting the values you would get the formula. I think this graphical way is a much interesting way of looking at the formula as algebra here can be too abstract for people new to it.






                                                                              share|cite|improve this answer











                                                                              $endgroup$













                                                                              • $begingroup$
                                                                                It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                                $endgroup$
                                                                                – David K
                                                                                Mar 8 '16 at 16:03
















                                                                              3












                                                                              $begingroup$

                                                                              Graphical Approach
                                                                              The following method actually requires some knowledge about the graph of parabola and calculus. This is just another way to view the formula.
                                                                              For $a>0,,
                                                                              f^{prime}(x)=0,left[x=-frac{b}{2a}right]$ would give you the minima. Now substitute this in the expression to see whether it attains any negative value. If it is positive here then it wont have any real roots for obvious reasons given that you have sufficient knowledge about functions.
                                                                              Now if the function has real roots then the arithmetic mean of the roots would give the axis of parabola. The roots are symmetrical about this line so if we somehow find the difference between the roots and add half of it to $-frac{b}{2a}$ then we would get one root and subtracting half of difference would give the other root.We know the sum of roots is $-frac{b}{a}$ and product of roots is $frac{c}{a}$. So
                                                                              begin{equation}
                                                                              F+G=-frac{a}{b},,FG=frac{c}{a}$
                                                                              end{equation}
                                                                              therefore, $(F-G)^2 = left(-frac{a}{b}right)^2 -4frac{c}{a}$
                                                                              taking root,
                                                                              $|F-G|=left[left(-frac{a}{b}right)^2 -4frac{c}{a}right]^{frac{1}{2}}$
                                                                              therefore the roots are
                                                                              $-frac{b}{a} + frac{|F-G|}{2},-frac{b}{a} - frac{|F-G|}{2}$ enter image description here
                                                                              Substituting the values you would get the formula. I think this graphical way is a much interesting way of looking at the formula as algebra here can be too abstract for people new to it.






                                                                              share|cite|improve this answer











                                                                              $endgroup$













                                                                              • $begingroup$
                                                                                It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                                $endgroup$
                                                                                – David K
                                                                                Mar 8 '16 at 16:03














                                                                              3












                                                                              3








                                                                              3





                                                                              $begingroup$

                                                                              Graphical Approach
                                                                              The following method actually requires some knowledge about the graph of parabola and calculus. This is just another way to view the formula.
                                                                              For $a>0,,
                                                                              f^{prime}(x)=0,left[x=-frac{b}{2a}right]$ would give you the minima. Now substitute this in the expression to see whether it attains any negative value. If it is positive here then it wont have any real roots for obvious reasons given that you have sufficient knowledge about functions.
                                                                              Now if the function has real roots then the arithmetic mean of the roots would give the axis of parabola. The roots are symmetrical about this line so if we somehow find the difference between the roots and add half of it to $-frac{b}{2a}$ then we would get one root and subtracting half of difference would give the other root.We know the sum of roots is $-frac{b}{a}$ and product of roots is $frac{c}{a}$. So
                                                                              begin{equation}
                                                                              F+G=-frac{a}{b},,FG=frac{c}{a}$
                                                                              end{equation}
                                                                              therefore, $(F-G)^2 = left(-frac{a}{b}right)^2 -4frac{c}{a}$
                                                                              taking root,
                                                                              $|F-G|=left[left(-frac{a}{b}right)^2 -4frac{c}{a}right]^{frac{1}{2}}$
                                                                              therefore the roots are
                                                                              $-frac{b}{a} + frac{|F-G|}{2},-frac{b}{a} - frac{|F-G|}{2}$ enter image description here
                                                                              Substituting the values you would get the formula. I think this graphical way is a much interesting way of looking at the formula as algebra here can be too abstract for people new to it.






                                                                              share|cite|improve this answer











                                                                              $endgroup$



                                                                              Graphical Approach
                                                                              The following method actually requires some knowledge about the graph of parabola and calculus. This is just another way to view the formula.
                                                                              For $a>0,,
                                                                              f^{prime}(x)=0,left[x=-frac{b}{2a}right]$ would give you the minima. Now substitute this in the expression to see whether it attains any negative value. If it is positive here then it wont have any real roots for obvious reasons given that you have sufficient knowledge about functions.
                                                                              Now if the function has real roots then the arithmetic mean of the roots would give the axis of parabola. The roots are symmetrical about this line so if we somehow find the difference between the roots and add half of it to $-frac{b}{2a}$ then we would get one root and subtracting half of difference would give the other root.We know the sum of roots is $-frac{b}{a}$ and product of roots is $frac{c}{a}$. So
                                                                              begin{equation}
                                                                              F+G=-frac{a}{b},,FG=frac{c}{a}$
                                                                              end{equation}
                                                                              therefore, $(F-G)^2 = left(-frac{a}{b}right)^2 -4frac{c}{a}$
                                                                              taking root,
                                                                              $|F-G|=left[left(-frac{a}{b}right)^2 -4frac{c}{a}right]^{frac{1}{2}}$
                                                                              therefore the roots are
                                                                              $-frac{b}{a} + frac{|F-G|}{2},-frac{b}{a} - frac{|F-G|}{2}$ enter image description here
                                                                              Substituting the values you would get the formula. I think this graphical way is a much interesting way of looking at the formula as algebra here can be too abstract for people new to it.







                                                                              share|cite|improve this answer














                                                                              share|cite|improve this answer



                                                                              share|cite|improve this answer








                                                                              edited Mar 8 '16 at 15:54









                                                                              Armando j18eos

                                                                              2,63511328




                                                                              2,63511328










                                                                              answered Mar 8 '16 at 15:20









                                                                              Sea Men on dutySea Men on duty

                                                                              562




                                                                              562












                                                                              • $begingroup$
                                                                                It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                                $endgroup$
                                                                                – David K
                                                                                Mar 8 '16 at 16:03


















                                                                              • $begingroup$
                                                                                It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                                $endgroup$
                                                                                – David K
                                                                                Mar 8 '16 at 16:03
















                                                                              $begingroup$
                                                                              It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                              $endgroup$
                                                                              – David K
                                                                              Mar 8 '16 at 16:03




                                                                              $begingroup$
                                                                              It's interesting that because the "algebra here can be too abstract for people new to it", we introduce calculus instead. But it appears (graphically) that for $a>0$, the minimum of $ax^2+bx+c$ is at $x_0=-b/(2a)$ (with a factor of $2$ in the divisor, don't forget that!) and you can prove this; see here for example, although that proof uses algebra again, so you might not like it.
                                                                              $endgroup$
                                                                              – David K
                                                                              Mar 8 '16 at 16:03











                                                                              2












                                                                              $begingroup$

                                                                              Our approach exploits the fact that any differentiable function $f:mathbb{R} rightarrow mathbb{R}$ is uniquely specified by its derivative and a point on its graph. The quadratic polynomial $P(x) = ax^2 + bx + c$ where $a neq 0$ satisfies $P(0) = c$ and $P'(x) = 2ax + b$.



                                                                              We have $P'(x) = 2ax + b = a(2(x+frac {b}{2a}))$. We can deduce from this that $g(x) = acdot(x+frac{b}{2a})^2 + C$ has the same derivative as $P(x)$. To make it the same function, it must have the same value at a point on its graph, say at $x=0$.



                                                                              In other words, we must make $g(0) = frac{b^2}{4a} + C = c$ which can be done if we let $C = c - frac{b^2}{4a}$. We have thus shown $ax^2 + bx + c = acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a}$



                                                                              To derive the quadratic formula, $acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a} = 0 $. This can be solved using elementary algebraic manipulations, from which we get the desired $x = frac{-bpm sqrt{b^{2} - 4 ac}}{2a}$






                                                                              share|cite|improve this answer









                                                                              $endgroup$


















                                                                                2












                                                                                $begingroup$

                                                                                Our approach exploits the fact that any differentiable function $f:mathbb{R} rightarrow mathbb{R}$ is uniquely specified by its derivative and a point on its graph. The quadratic polynomial $P(x) = ax^2 + bx + c$ where $a neq 0$ satisfies $P(0) = c$ and $P'(x) = 2ax + b$.



                                                                                We have $P'(x) = 2ax + b = a(2(x+frac {b}{2a}))$. We can deduce from this that $g(x) = acdot(x+frac{b}{2a})^2 + C$ has the same derivative as $P(x)$. To make it the same function, it must have the same value at a point on its graph, say at $x=0$.



                                                                                In other words, we must make $g(0) = frac{b^2}{4a} + C = c$ which can be done if we let $C = c - frac{b^2}{4a}$. We have thus shown $ax^2 + bx + c = acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a}$



                                                                                To derive the quadratic formula, $acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a} = 0 $. This can be solved using elementary algebraic manipulations, from which we get the desired $x = frac{-bpm sqrt{b^{2} - 4 ac}}{2a}$






                                                                                share|cite|improve this answer









                                                                                $endgroup$
















                                                                                  2












                                                                                  2








                                                                                  2





                                                                                  $begingroup$

                                                                                  Our approach exploits the fact that any differentiable function $f:mathbb{R} rightarrow mathbb{R}$ is uniquely specified by its derivative and a point on its graph. The quadratic polynomial $P(x) = ax^2 + bx + c$ where $a neq 0$ satisfies $P(0) = c$ and $P'(x) = 2ax + b$.



                                                                                  We have $P'(x) = 2ax + b = a(2(x+frac {b}{2a}))$. We can deduce from this that $g(x) = acdot(x+frac{b}{2a})^2 + C$ has the same derivative as $P(x)$. To make it the same function, it must have the same value at a point on its graph, say at $x=0$.



                                                                                  In other words, we must make $g(0) = frac{b^2}{4a} + C = c$ which can be done if we let $C = c - frac{b^2}{4a}$. We have thus shown $ax^2 + bx + c = acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a}$



                                                                                  To derive the quadratic formula, $acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a} = 0 $. This can be solved using elementary algebraic manipulations, from which we get the desired $x = frac{-bpm sqrt{b^{2} - 4 ac}}{2a}$






                                                                                  share|cite|improve this answer









                                                                                  $endgroup$



                                                                                  Our approach exploits the fact that any differentiable function $f:mathbb{R} rightarrow mathbb{R}$ is uniquely specified by its derivative and a point on its graph. The quadratic polynomial $P(x) = ax^2 + bx + c$ where $a neq 0$ satisfies $P(0) = c$ and $P'(x) = 2ax + b$.



                                                                                  We have $P'(x) = 2ax + b = a(2(x+frac {b}{2a}))$. We can deduce from this that $g(x) = acdot(x+frac{b}{2a})^2 + C$ has the same derivative as $P(x)$. To make it the same function, it must have the same value at a point on its graph, say at $x=0$.



                                                                                  In other words, we must make $g(0) = frac{b^2}{4a} + C = c$ which can be done if we let $C = c - frac{b^2}{4a}$. We have thus shown $ax^2 + bx + c = acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a}$



                                                                                  To derive the quadratic formula, $acdot(x+frac{b}{2a})^2 + c - frac{b^2}{4a} = 0 $. This can be solved using elementary algebraic manipulations, from which we get the desired $x = frac{-bpm sqrt{b^{2} - 4 ac}}{2a}$







                                                                                  share|cite|improve this answer












                                                                                  share|cite|improve this answer



                                                                                  share|cite|improve this answer










                                                                                  answered Feb 19 '16 at 23:06









                                                                                  MathematicsStudent1122MathematicsStudent1122

                                                                                  8,62622467




                                                                                  8,62622467























                                                                                      1












                                                                                      $begingroup$

                                                                                      If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant is called completing the square). The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.



                                                                                      So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B pm sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.






                                                                                      share|cite|improve this answer









                                                                                      $endgroup$


















                                                                                        1












                                                                                        $begingroup$

                                                                                        If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant is called completing the square). The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.



                                                                                        So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B pm sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.






                                                                                        share|cite|improve this answer









                                                                                        $endgroup$
















                                                                                          1












                                                                                          1








                                                                                          1





                                                                                          $begingroup$

                                                                                          If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant is called completing the square). The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.



                                                                                          So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B pm sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.






                                                                                          share|cite|improve this answer









                                                                                          $endgroup$



                                                                                          If one is interested only in the roots one can concentrate on the monic equations. Now any monic quadratic expression becomes the square of a linear expression by addition of a suitable constant (finding what the suitable constant is called completing the square). The additive constant is uniformly the same expression in the coefficients of the given quadratic equation.



                                                                                          So the given equation has the same solutions as the modified equation of the form $(x+B)^2=A$. SO the formula for the roots are of the form $x=-B pm sqrt A$. Thats the reason all quadratic equations have solutions expressed in a uniform manner.







                                                                                          share|cite|improve this answer












                                                                                          share|cite|improve this answer



                                                                                          share|cite|improve this answer










                                                                                          answered Jul 19 '16 at 8:49









                                                                                          P VanchinathanP Vanchinathan

                                                                                          15.1k12136




                                                                                          15.1k12136























                                                                                              0












                                                                                              $begingroup$

                                                                                              There are essentially two questions being asked here:
                                                                                              1. Why does the quadratic formula apply to ALL quadratic equations?
                                                                                              2. Why is the quadratic formula correct / How can I prove this?



                                                                                              Most of the answers have focussed on the second version of this question, and more specifically, at simply coming up with different proofs, rather than instructing how to come up with one yourself. Fair enough, since these proofs have been incredibly instructive to many readers including me.



                                                                                              But it does leave a bit of a gap. I'll address the first part of that gap by answering the first question: why does the quadratic formula actually apply to all quadratic equations?



                                                                                              That's because the quadratic formula expresses its solution without assuming any information about the actual value of any of the constants a, b, and c. Therefore, it can be applied to a quadratic equation with any values for these three constants.



                                                                                              This is what mathematics is all about: finding general solutions to things so that you can use the same solution method for a wide array of problems.



                                                                                              The same could be said about why the number "3" can be used to describe any group of that size; a group of three sheep, three people, three bricks, whatever. The whole reason why that concept was invented and why it survived the passage of time is because it's so useful.



                                                                                              It's useful because it can be applied to a wide array of different things, rather than requiring a re-invention of the wheel for every different kind of object. This can, of course, only be done by presuming as little as possible to whatever the concept can be applied to.



                                                                                              And that's why concepts like the quadratic formula show up in your maths classes in the first place. They solve a very important problem because many real-life things can be modeled as polynomials, especially the lower ones like the linear and quadratic, so being able to solve such a thing with just one "wheel" is incredibly important and useful.



                                                                                              Hopefully that adds a little perspective on the general applicability and consequent utility of the quadratic formula, the former essentially being what the question was inquiring under its most literal interpretation.






                                                                                              share|cite|improve this answer









                                                                                              $endgroup$


















                                                                                                0












                                                                                                $begingroup$

                                                                                                There are essentially two questions being asked here:
                                                                                                1. Why does the quadratic formula apply to ALL quadratic equations?
                                                                                                2. Why is the quadratic formula correct / How can I prove this?



                                                                                                Most of the answers have focussed on the second version of this question, and more specifically, at simply coming up with different proofs, rather than instructing how to come up with one yourself. Fair enough, since these proofs have been incredibly instructive to many readers including me.



                                                                                                But it does leave a bit of a gap. I'll address the first part of that gap by answering the first question: why does the quadratic formula actually apply to all quadratic equations?



                                                                                                That's because the quadratic formula expresses its solution without assuming any information about the actual value of any of the constants a, b, and c. Therefore, it can be applied to a quadratic equation with any values for these three constants.



                                                                                                This is what mathematics is all about: finding general solutions to things so that you can use the same solution method for a wide array of problems.



                                                                                                The same could be said about why the number "3" can be used to describe any group of that size; a group of three sheep, three people, three bricks, whatever. The whole reason why that concept was invented and why it survived the passage of time is because it's so useful.



                                                                                                It's useful because it can be applied to a wide array of different things, rather than requiring a re-invention of the wheel for every different kind of object. This can, of course, only be done by presuming as little as possible to whatever the concept can be applied to.



                                                                                                And that's why concepts like the quadratic formula show up in your maths classes in the first place. They solve a very important problem because many real-life things can be modeled as polynomials, especially the lower ones like the linear and quadratic, so being able to solve such a thing with just one "wheel" is incredibly important and useful.



                                                                                                Hopefully that adds a little perspective on the general applicability and consequent utility of the quadratic formula, the former essentially being what the question was inquiring under its most literal interpretation.






                                                                                                share|cite|improve this answer









                                                                                                $endgroup$
















                                                                                                  0












                                                                                                  0








                                                                                                  0





                                                                                                  $begingroup$

                                                                                                  There are essentially two questions being asked here:
                                                                                                  1. Why does the quadratic formula apply to ALL quadratic equations?
                                                                                                  2. Why is the quadratic formula correct / How can I prove this?



                                                                                                  Most of the answers have focussed on the second version of this question, and more specifically, at simply coming up with different proofs, rather than instructing how to come up with one yourself. Fair enough, since these proofs have been incredibly instructive to many readers including me.



                                                                                                  But it does leave a bit of a gap. I'll address the first part of that gap by answering the first question: why does the quadratic formula actually apply to all quadratic equations?



                                                                                                  That's because the quadratic formula expresses its solution without assuming any information about the actual value of any of the constants a, b, and c. Therefore, it can be applied to a quadratic equation with any values for these three constants.



                                                                                                  This is what mathematics is all about: finding general solutions to things so that you can use the same solution method for a wide array of problems.



                                                                                                  The same could be said about why the number "3" can be used to describe any group of that size; a group of three sheep, three people, three bricks, whatever. The whole reason why that concept was invented and why it survived the passage of time is because it's so useful.



                                                                                                  It's useful because it can be applied to a wide array of different things, rather than requiring a re-invention of the wheel for every different kind of object. This can, of course, only be done by presuming as little as possible to whatever the concept can be applied to.



                                                                                                  And that's why concepts like the quadratic formula show up in your maths classes in the first place. They solve a very important problem because many real-life things can be modeled as polynomials, especially the lower ones like the linear and quadratic, so being able to solve such a thing with just one "wheel" is incredibly important and useful.



                                                                                                  Hopefully that adds a little perspective on the general applicability and consequent utility of the quadratic formula, the former essentially being what the question was inquiring under its most literal interpretation.






                                                                                                  share|cite|improve this answer









                                                                                                  $endgroup$



                                                                                                  There are essentially two questions being asked here:
                                                                                                  1. Why does the quadratic formula apply to ALL quadratic equations?
                                                                                                  2. Why is the quadratic formula correct / How can I prove this?



                                                                                                  Most of the answers have focussed on the second version of this question, and more specifically, at simply coming up with different proofs, rather than instructing how to come up with one yourself. Fair enough, since these proofs have been incredibly instructive to many readers including me.



                                                                                                  But it does leave a bit of a gap. I'll address the first part of that gap by answering the first question: why does the quadratic formula actually apply to all quadratic equations?



                                                                                                  That's because the quadratic formula expresses its solution without assuming any information about the actual value of any of the constants a, b, and c. Therefore, it can be applied to a quadratic equation with any values for these three constants.



                                                                                                  This is what mathematics is all about: finding general solutions to things so that you can use the same solution method for a wide array of problems.



                                                                                                  The same could be said about why the number "3" can be used to describe any group of that size; a group of three sheep, three people, three bricks, whatever. The whole reason why that concept was invented and why it survived the passage of time is because it's so useful.



                                                                                                  It's useful because it can be applied to a wide array of different things, rather than requiring a re-invention of the wheel for every different kind of object. This can, of course, only be done by presuming as little as possible to whatever the concept can be applied to.



                                                                                                  And that's why concepts like the quadratic formula show up in your maths classes in the first place. They solve a very important problem because many real-life things can be modeled as polynomials, especially the lower ones like the linear and quadratic, so being able to solve such a thing with just one "wheel" is incredibly important and useful.



                                                                                                  Hopefully that adds a little perspective on the general applicability and consequent utility of the quadratic formula, the former essentially being what the question was inquiring under its most literal interpretation.







                                                                                                  share|cite|improve this answer












                                                                                                  share|cite|improve this answer



                                                                                                  share|cite|improve this answer










                                                                                                  answered Sep 5 '17 at 19:43









                                                                                                  RalphDRalphD

                                                                                                  345




                                                                                                  345

















                                                                                                      protected by Zev Chonoles Jan 19 '16 at 18:14



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