Find range of values for 'k' that gives this equation 2 distinct real roots?
$begingroup$
I stuck on a question which is asking me to find the range of values for k.
the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots.
$$3x^2+kx+2=0$$
this is what i have done so far:
$$b^2-4ac>0$$
Note - since there are two distinct real roots
$$k^2-24>0$$
considering $$k^2-24=0$$
$$k = +-sqrt{24}$$
$$k = 2*sqrt{6} $$or
$$k = -2*sqrt{6}$$
my question is how do I work out to range of values for k which makes the inequality true.
Can you solve the inequality step-by-step $$k^2-24>0$$
inequality quadratics
asked Mar 5 '15 at 0:02
user133745user133745
39117
$endgroup$
add a comment |
$begingroup$
I stuck on a question which is asking me to find the range of values for k.
the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots.
$$3x^2+kx+2=0$$
this is what i have done so far:
$$b^2-4ac>0$$
Note - since there are two distinct real roots
$$k^2-24>0$$
considering $$k^2-24=0$$
$$k = +-sqrt{24}$$
$$k = 2*sqrt{6} $$or
$$k = -2*sqrt{6}$$
my question is how do I work out to range of values for k which makes the inequality true.
Can you solve the inequality step-by-step $$k^2-24>0$$
inequality quadratics
asked Mar 5 '15 at 0:02
user133745user133745
39117
$endgroup$
add a comment |
$begingroup$
I stuck on a question which is asking me to find the range of values for k.
the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots.
$$3x^2+kx+2=0$$
this is what i have done so far:
$$b^2-4ac>0$$
Note - since there are two distinct real roots
$$k^2-24>0$$
considering $$k^2-24=0$$
$$k = +-sqrt{24}$$
$$k = 2*sqrt{6} $$or
$$k = -2*sqrt{6}$$
my question is how do I work out to range of values for k which makes the inequality true.
Can you solve the inequality step-by-step $$k^2-24>0$$
inequality quadratics
asked Mar 5 '15 at 0:02
user133745user133745
39117
$endgroup$
I stuck on a question which is asking me to find the range of values for k.
the question is : By considering the discriminant, or otherwise, find the range of values of 'k' that gives the equation 2 distinct roots.
$$3x^2+kx+2=0$$
this is what i have done so far:
$$b^2-4ac>0$$
Note - since there are two distinct real roots
$$k^2-24>0$$
considering $$k^2-24=0$$
$$k = +-sqrt{24}$$
$$k = 2*sqrt{6} $$or
$$k = -2*sqrt{6}$$
my question is how do I work out to range of values for k which makes the inequality true.
Can you solve the inequality step-by-step $$k^2-24>0$$
inequality quadratics
inequality quadratics
asked Mar 5 '15 at 0:02
user133745user133745
39117
asked Mar 5 '15 at 0:02
user133745user133745
39117
edited Mar 5 '15 at 0:35
user133745
asked Mar 5 '15 at 0:02
user133745user133745
39117
asked Mar 5 '15 at 0:02
user133745user133745
39117
asked Mar 5 '15 at 0:02
user133745user133745
39117
39117
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > sqrt{24} = 2sqrt{6}$$
recalling that $sqrt{x^2} = |x|$ ($x in Bbb R$). We can then write this as
$$k > 2sqrt{6} text{or} k < -2sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_pm$, say, then you just need to test the intervals $(-infty, k_-)$, $(k_-,k_+)$ and $(k_+,infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_pm = pm 2sqrt{6}$.)
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
$endgroup$
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > sqrt{24} = 2sqrt{6}$$
recalling that $sqrt{x^2} = |x|$ ($x in Bbb R$). We can then write this as
$$k > 2sqrt{6} text{or} k < -2sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_pm$, say, then you just need to test the intervals $(-infty, k_-)$, $(k_-,k_+)$ and $(k_+,infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_pm = pm 2sqrt{6}$.)
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
$endgroup$
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
add a comment |
$begingroup$
You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > sqrt{24} = 2sqrt{6}$$
recalling that $sqrt{x^2} = |x|$ ($x in Bbb R$). We can then write this as
$$k > 2sqrt{6} text{or} k < -2sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_pm$, say, then you just need to test the intervals $(-infty, k_-)$, $(k_-,k_+)$ and $(k_+,infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_pm = pm 2sqrt{6}$.)
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
$endgroup$
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
add a comment |
$begingroup$
You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > sqrt{24} = 2sqrt{6}$$
recalling that $sqrt{x^2} = |x|$ ($x in Bbb R$). We can then write this as
$$k > 2sqrt{6} text{or} k < -2sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_pm$, say, then you just need to test the intervals $(-infty, k_-)$, $(k_-,k_+)$ and $(k_+,infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_pm = pm 2sqrt{6}$.)
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
$endgroup$
You are almost there. You get, correctly, that there are two distinct, real roots if and only if $k^2 - 24 > 0$. So this is true if and only if $k^2 > 24$. Since both sides are greater than $0$, we may apply the (positive) square root function to both sides to obtain the final inequality
$$|k| > sqrt{24} = 2sqrt{6}$$
recalling that $sqrt{x^2} = |x|$ ($x in Bbb R$). We can then write this as
$$k > 2sqrt{6} text{or} k < -2sqrt{6}.$$
Observe that these are the two critical points that you found. This is where the situation swaps from not having two, distinct real roots to having them. If you just set the inequality to be an equality and solve, then you find the critical points $k_pm$, say, then you just need to test the intervals $(-infty, k_-)$, $(k_-,k_+)$ and $(k_+,infty)$. Note that it's actually just enough to test one of these intervals, as the adjacent ones must be different to it: in particular, if you test some $k in (k_-,k_+)$ in your inequality and find that it does not satisfy, ie you don't have two real, distinct roots, then you know that for $k>k_+$ and $k<k_-$ you will. (In our case, $k_pm = pm 2sqrt{6}$.)
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
answered Mar 5 '15 at 0:06
Sam TSam T
4,0101032
4,0101032
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
add a comment |
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
I'm a little confused. how do you determine whether to use < or > sign after you get the 2 roots $$k=+-2sqrt{6}$$
$endgroup$
– user133745
Mar 5 '15 at 0:15
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
@user133745 Remember that $|k| > 2sqrt{6}$ means that $k$ is more than $2sqrt{6}$ units from $0$ on the real number line, which implies that $k > 2sqrt{6}$ or $k < -2sqrt{6}$.
$endgroup$
– N. F. Taussig
Mar 5 '15 at 1:00
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
$begingroup$
Exactly as @N.F.Taussig has said. :) - Alternatively, as I say at the end, you can simply test a value in one of the intervals: Test 0, giving $3x^2 + 2 = 0$; this doesn't have any real roots, so $0$ certainly isn't in your region. Hence the adjacent intervals are (as they are separated by the critical points that you found). The way N.F.Taussig said is much easier for this case, but if you had a quartic ($x^4 + ...$), for example, then the other way can be helpful. :)
$endgroup$
– Sam T
Mar 5 '15 at 8:43
add a comment |
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