How do you solve $2 x^{4} + x^{2} - 3 = 0$ $2x^4+x^2-3=0$ ?

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How do you solve $2 x^{4} + x^{2} - 3 = 0$ $2x^4+x^2-3=0$ ?

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Answer 1 · 2016-02-28T00:45:24

Simplify by treating as a quadratic in $x^{2}$ $x^2$ and factor by grouping.

Explanation:

This is a "quadratic in $x^{2}$ $x^2$ " which we can factor by grouping:

$2 x^{4} + x^{2} - 3$ $2x^4+x^2-3$

$= 2 {(x^{2})}^{2} + (x^{2}) - 3$ $=2(x^2)^2+(x^2)-3$

$= (2 {(x^{2})}^{2} - 2 (x^{2})) + (3 (x^{2}) - 3)$ $=(2(x^2)^2-2(x^2))+(3(x^2)-3)$

$= 2 x^{2} (x^{2} - 1) + 3 (x^{2} - 1)$ $=2x^2(x^2-1)+3(x^2-1)$

$= (2 x^{2} + 3) (x^{2} - 1)$ $=(2x^2+3)(x^2-1)$

$= (2 x^{2} + 3) (x - 1) (x + 1)$ $=(2x^2+3)(x-1)(x+1)$

The remaining quadratic factor has no linear factors with Real coefficients, but it can be factored if we allow Complex coefficients:

$= (\sqrt{2} x - \sqrt{3} i) (\sqrt{2} x + \sqrt{3} i) (x - 1) (x + 1)$ $=(sqrt(2)x-sqrt(3)i)(sqrt(2)x+sqrt(3)i)(x-1)(x+1)$

Hence the roots of the equation are:

$x = \pm 1$ $x = +-1$ and $x = \pm \frac{\sqrt{3}}{\sqrt{2}} i = \pm \frac{\sqrt{6}}{2} i$ $x = +-sqrt(3)/sqrt(2)i = +-sqrt(6)/2i$

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How do you solve $2 x^{4} + x^{2} - 3 = 0$ $2x^4+x^2-3=0$ ?

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