Quadratics substitute problem
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Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?
algebra-precalculus
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
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add a comment |
$begingroup$
Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?
algebra-precalculus
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
$endgroup$
2
$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19
add a comment |
$begingroup$
Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?
algebra-precalculus
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
$endgroup$
Ok, so I can make the substitution $z^4=u^2$ to solve $az^4+bz^2+c=0,:forall:a,b,cinmathbb{R},:aneq0$?
algebra-precalculus
algebra-precalculus
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
asked Dec 9 '18 at 22:06
NumbersNumbers
1116
1116
2
$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19
add a comment |
2
$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19
2
2
$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This is what is known as a biquadratic. Notice we can rewrite it as:
$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$
In other words, yes, what you are saying is correct.
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is what is known as a biquadratic. Notice we can rewrite it as:
$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$
In other words, yes, what you are saying is correct.
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
$endgroup$
add a comment |
$begingroup$
This is what is known as a biquadratic. Notice we can rewrite it as:
$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$
In other words, yes, what you are saying is correct.
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
$endgroup$
add a comment |
$begingroup$
This is what is known as a biquadratic. Notice we can rewrite it as:
$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$
In other words, yes, what you are saying is correct.
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
$endgroup$
This is what is known as a biquadratic. Notice we can rewrite it as:
$$a(z^2)^2+b(z^2)^1+c(z^2)^0=0$$ and we have a quadratic. So we use the quadratic equation for:
$$z^2=frac{-bpmsqrt{b^2-4ac}}{2a}to z=pmsqrt{frac{-bpmsqrt{b^2-4ac}}{2a}}$$
In other words, yes, what you are saying is correct.
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
answered Dec 9 '18 at 22:10
Rhys HughesRhys Hughes
6,7001530
6,7001530
add a comment |
add a comment |
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$begingroup$
Yes, you can, together with $z^2=u$.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:08
$begingroup$
ok, but then $z^2=|u|$ and I am losing solutions that way!
$endgroup$
– Numbers
Dec 9 '18 at 22:10
$begingroup$
@Numbers have a look at my answer. The two instances of $pm$ indicate there are four solutions, as expected of a fourth degree polynomial. Also you're losing values because instead of taking $z^2=|u|$ you should take $z^2=pm u$
$endgroup$
– Rhys Hughes
Dec 9 '18 at 22:14
$begingroup$
Yeah, u r right, thx man!
$endgroup$
– Numbers
Dec 9 '18 at 22:19