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

Feb 28, 2016

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

Explanation:

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

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

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

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

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

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

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

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

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

Hence the roots of the equation are:

$x = \pm 1$ and $x = \pm \frac{\sqrt{3}}{\sqrt{2}} i = \pm \frac{\sqrt{6}}{2} i$