# How do you find the complex roots of x^4-10x^2+9=0?

Jan 10, 2017

The roots are $\pm 3$ and $\pm 1$

#### Explanation:

Note that $9 + 1 = 10$ and $9 \cdot 1 = 9$.

Hence we find:

$0 = {x}^{4} - 10 {x}^{2} + 9$

$\textcolor{w h i t e}{0} = \left({x}^{2} - 9\right) \left({x}^{2} - 1\right)$

$\textcolor{w h i t e}{0} = \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {1}^{2}\right)$

$\textcolor{w h i t e}{0} = \left(x - 3\right) \left(x + 3\right) \left(x - 1\right) \left(x + 1\right)$

So the four roots of this quartic equation are $\pm 3$ and $\pm 1$

These are Real numbers, but any Real number is also a Complex number (of the form $a + 0 i$ if you wish).