# How do you find all the real and complex roots of x^4-6x^2+5=0?

Jan 22, 2016

$x = - \sqrt{5} , - 1 , 1 , \sqrt{5}$

#### Explanation:

Notice that this is very similar to ${x}^{2} - 6 x + 5$, which could be factored into $\left(x - 5\right) \left(x - 1\right)$. Similarly, this polynomial can be factored as

$\left({x}^{2} - 5\right) \left({x}^{2} - 1\right) = 0$

To solve this, set each individual part being multiplied equal to $0$ and solve.

${x}^{2} - 5 = 0$
${x}^{2} = 5$
$x = \pm \sqrt{5}$

${x}^{2} - 1 = 0$
${x}^{2} = 1$
$x = \pm 1$

These are the equation's four roots (where the equation crosses the $x$ axis). We can check a graph:

graph{x^4-6x^2+5 [-19.53, 21.01, -6, 14.28]}

Note that $\sqrt{5} \approx 2.24$.