Question #9f7e1

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Question #9f7e1

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Answer 1 · 2017-12-22T10:54:22

$x = \frac{\sqrt{2}}{2} or x = - \frac{\sqrt{2}}{2}$ $x=sqrt2/2 or x=-sqrt2/2$
$s o l u t i o n s e t = {\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}}$ $solution set= {sqrt2/2,-sqrt2/2}$

Explanation:

$2 x^{4} + 5 x^{2} - 3 = 0$ $2x^4+5x^2-3=0$
let $y = x^{2}$ $y=x^2$
so, $y^{2} = x^{4}$ $y^2=x^4$

$2 x^{4} + 5 x^{2} - 3 = 0$ $2x^4+5x^2-3=0$
$2 y^{2} + 5 y - 3 = 0$ $2y^2+5y-3=0$
$(2 y - 1) (y + 3) = 0$ $(2y-1)(y+3)=0$

There are two cases here:

Case 1:
$2 y - 1 = 0$ $2y-1=0$
$y = \frac{1}{2}$ $y=1/2$
$x^{2} = \frac{1}{2}$ $x^2=1/2$
$x = \pm \frac{1}{\sqrt{2}}$ $x=+-1/sqrt2$
$x = \pm \frac{\sqrt{2}}{2}$ $x=+-sqrt2/2$

Case 2:
$y + 3 = 0$ $y+3=0$
$y = - 3$ $y=-3$
$x^{2} = - 3$ $x^2=-3$
$because x^2$ is never negative
$:. x^2=-3$ is impossible
$:.$ Unsuitable

$:.$ $solution set= {sqrt2/2,-sqrt2/2}$

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Question #9f7e1

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