# How do you find the complex roots of a^4+a^2-2=0?

Nov 24, 2016

This quartic has Real roots $a = \pm 1$ and Complex roots $a = \pm \sqrt{2} i$

#### Explanation:

We can write this quartic as a quadratic in ${a}^{2}$:

${\left({a}^{2}\right)}^{2} + \left({a}^{2}\right) - 2 = 0$

Note that the sum of the coefficients is $0$. That is:

$1 + 1 - 2 = 0$

Hence ${a}^{2} = 1$ is a solution and $\left({a}^{2} - 1\right)$ a factor:

$0 = {\left({a}^{2}\right)}^{2} + \left({a}^{2}\right) - 2 = \left({a}^{2} - 1\right) \left({a}^{2} + 2\right)$

We can factor this into linear factors with Real or imaginary coefficients using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

as follows:

$\left({a}^{2} - 1\right) \left({a}^{2} + 2\right) = \left({a}^{2} - {1}^{2}\right) \left({a}^{2} - {\left(\sqrt{2} i\right)}^{2}\right)$

$\textcolor{w h i t e}{\left({a}^{2} - 1\right) \left({a}^{2} + 2\right)} = \left(a - 1\right) \left(a + 1\right) \left(a - \sqrt{2} i\right) \left(a + \sqrt{2} i\right)$

So there are Real roots: $a = \pm 1$ and Complex roots: $a = \pm \sqrt{2} i$