# How do you find the roots of x^4-5x^2+4=0?

Mar 22, 2018

$x = \pm 2 , \pm 1$

#### Explanation:

Let $y = {x}^{2}$.

${y}^{2} - 5 y + 4 = 0$

This is a nice and simple trinomial to factor.

$\left(y - 4\right) \left(y - 1\right) = 0$

$y = 4 \mathmr{and} y = 1$

We now reverse our substitution.

${x}^{2} = 4 \mathmr{and} {x}^{2} = 1$

$x = \pm 2 \mathmr{and} \pm 1$

The graph of the function $f \left(x\right) = {x}^{4} - 5 {x}^{2} + 4$ confirms that there are zeroes at $x = \pm 2 \mathmr{and} \pm 1$.

graph{x^4 - 5x^2 + 4 [-10, 10, -5, 5]}

Hopefully this helps!