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 March 9th, 2019, 01:14 AM #1 Member   Joined: Aug 2018 From: România Posts: 85 Thanks: 6 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. All the best, Integrator
 March 9th, 2019, 02:56 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra \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} Thanks from topsquark
March 9th, 2019, 09:12 PM   #3
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Quote:
 Originally Posted by v8archie \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

 March 9th, 2019, 09:45 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 $\sqrt[3]{x^4-6x^2+9}-3a\sqrt[3]{x^2-3}+2a^2 = 0$ $\sqrt[3]{(x^2-3)^2}-3a\sqrt[3]{x^2-3}+2a^2 = 0$ 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. $x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$ Thanks from topsquark
March 10th, 2019, 09:36 AM   #5
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Quote:
 Originally Posted by Integrator 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?

March 10th, 2019, 09:33 PM   #6
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Quote:
 Originally Posted by romsek $\sqrt[3]{x^4-6x^2+9}-3a\sqrt[3]{x^2-3}+2a^2 = 0$ $\sqrt[3]{(x^2-3)^2}-3a\sqrt[3]{x^2-3}+2a^2 = 0$ 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,

Integrator

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