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Problem Of The Week #372 Jun 25th, 2019

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anemone

MHB POTW Director
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Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Thanks to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. MegaMoh
4. Greg
5. kaliprasad

Solution from Olinguito :
Let $y=\dfrac1{x^2}$; then
$$4x^6-6x^2+2\sqrt2\ =\ 0$$
$\implies\ \dfrac4{y^3}-\dfrac6y+2\sqrt2\ =\ 0$

$\implies\ 2\sqrt2y^3-6y^2+4\ =\ 0$

$\implies\ z^3-3z^2+4\ =\ 0$ where $z=\sqrt2y$

$\implies\ (z-2)^2(z+1)\ =\ 0$

$\implies\ z\ =\ -1,\,2$

$\implies\ y\ =\ \dfrac z{\sqrt2}\ =\ -\dfrac1{\sqrt2},\,\sqrt2$

$\implies\ x^2\ =\ \dfrac1y\ =\ -\sqrt2,\,\dfrac1{\sqrt2}$

$\implies\ \boxed{x\ =\ \pm i\sqrt[4]2,\,\pm\dfrac1{\sqrt[4]2}}$.

(If only real solutions are required, then $\boxed{x\ =\ \pm\dfrac1{\sqrt[4]2}}$.)
 
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