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March 9th, 2019, 01:14 AM   #1
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Re: An equation

Hello all,

Determine the set of all values of the parameter $\displaystyle a\geq 0$ knowing that the equation $\displaystyle \sqrt[3]{x^4-6x^2+9}-3a\sqrt[3]{x^2-3}+2a^2=0$ has all the real roots.

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March 9th, 2019, 02:56 AM   #2
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\begin{align}
a+b&=c \\
(a+b)^3 &= c^3 \\
a^3 + a^2b + ab^2 + b^3 &= c^3 \\
a^3 + (a+b)ab + b^3 &= c^3 \\
a^3 +abc + b^3 &= c^3
\end{align}
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March 9th, 2019, 09:12 PM   #3
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Quote:
Originally Posted by v8archie View Post
\begin{align}
a+b&=c \\
(a+b)^3 &= c^3 \\
a^3 + a^2b + ab^2 + b^3 &= c^3 \\
a^3 + (a+b)ab + b^3 &= c^3 \\
a^3 +abc + b^3 &= c^3
\end{align}
Hello,

Are you kidding?What is unclear?The problem I posted is from another forum.

All the best,

Integrator
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March 9th, 2019, 09:45 PM   #4
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$\sqrt[3]{x^4-6x^2+9}-3a\sqrt[3]{x^2-3}+2a^2 = 0$

$\sqrt[3]{(x^2-3)^2}-3a\sqrt[3]{x^2-3}+2a^2 = 0$

let $u = \sqrt[3]{x^2-3}$

$u^2 - 3au + 2a^2 = 0$

The discriminant

$D = (-3a)^2 - 4(2a^2) = a^2$

$\forall a \geq 0,~D\geq 0$

Thus the equation will always have real roots.

$x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$
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March 10th, 2019, 09:36 AM   #5
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Quote:
Originally Posted by Integrator View Post
Are you kidding?What is unclear?The problem I posted is from another forum.
So what? The suggestion I made came from my own head. If you want the solution, why not look on the other forum?
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March 10th, 2019, 09:33 PM   #6
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Quote:
Originally Posted by romsek View Post
$\sqrt[3]{x^4-6x^2+9}-3a\sqrt[3]{x^2-3}+2a^2 = 0$

$\sqrt[3]{(x^2-3)^2}-3a\sqrt[3]{x^2-3}+2a^2 = 0$

let $u = \sqrt[3]{x^2-3}$

$u^2 - 3au + 2a^2 = 0$

The discriminant

$D = (-3a)^2 - 4(2a^2) = a^2$

$\forall a \geq 0,~D\geq 0$

Thus the equation will always have real roots.

$x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$
Hello,

Some say that the equation has all the real roots only if $\displaystyle a=0$ , because the equation has $\displaystyle 12$ roots of which $\displaystyle 4$ are real and $\displaystyle 8$ complex roots. What are the complex roots?

All the best,

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