# How do you solve the quadratic with complex numbers given x^4+16x^2-225=0?

Nov 5, 2016

This quartic equation has roots $\pm 3$ and $\pm 5 i$

#### Explanation:

The difference of squares identity can be written:

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

We will use this several times.

$\textcolor{w h i t e}{}$
Given:

${x}^{4} + 16 {x}^{2} - 225 = 0$

This is a quartic in ${x}^{4}$, but effectively a quadratic in ${x}^{2}$. So we can make a start by treating it as such and using whatever method we like to solve the quadratic:

$0 = {x}^{4} + 16 {x}^{2} - 225$

$\textcolor{w h i t e}{0} = {\left({x}^{2}\right)}^{2} + 16 \left({x}^{2}\right) + 64 - 289$

$\textcolor{w h i t e}{0} = {\left({x}^{2} + 8\right)}^{2} - {17}^{2}$

$\textcolor{w h i t e}{0} = \left(\left({x}^{2} + 8\right) - 17\right) \left(\left({x}^{2} + 8\right) + 17\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - 9\right) \left({x}^{2} + 25\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(5 i\right)}^{2}\right)$

$\textcolor{w h i t e}{0} = \left(x - 3\right) \left(x + 3\right) \left(x - 5 i\right) \left(x + 5 i\right)$

Hence roots:

$x = \pm 3$ and $x = \pm 5 i$