# How do you solve x^4 – 8x^2 – 9 = 0?

Feb 11, 2016

x can be 3, -3, i, or -i.

#### Explanation:

If you can't find the factoring by looking at this, simplify the equation.

let ${x}^{2} = y$ to make things easier to see

Now we have ${y}^{2} - 8 y - 9 = 0$

See it now?

We can factor into $\left(y - 9\right) \left(y + 1\right)$

Now substitute back in ${x}^{2}$ for $y$.

$\left({x}^{2} - 9\right) \left({x}^{2} + 1\right)$

Since $\left({x}^{2} - 9\right)$ is a difference of two squares,

$\left(x - 3\right) \left(x + 3\right)$

Now we have $\left(x - 3\right) \left(x + 3\right) \left({x}^{2} + 1\right)$
x can be 3, -3 for the first two parts

${x}^{2} + 1 = 0$ can become ${x}^{2} = - 1$
Taking the positive and negative root means $x = \pm \sqrt{- 1}$

Thus $x = \pm i$ in addition to 3 and -3.