# How do you solve x^4 + x^2 = 1 ?

Mar 14, 2016

Solve as a quadratic in ${x}^{2}$ using the quadratic formula, then take square roots...

#### Explanation:

${x}^{4} + {x}^{2} = 1$

Subtract $1$ from both sides to get:

${x}^{4} + {x}^{2} - 1 = 0$

Writing ${x}^{4} = {\left({x}^{2}\right)}^{2}$ we have:

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

This is in the form $a {X}^{2} + b X + c = 0$ with $X = {x}^{2}$, $a = 1$, $b = 1$ and $c = - 1$.

We can use the quadratic formula to find:

${x}^{2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \cdot 1 \cdot - 1\right)}}{2 \cdot 1}$

$= \frac{- 1 \pm \sqrt{5}}{2}$

So:

$x = \pm \sqrt{\frac{- 1 + \sqrt{5}}{2}}$

Or:

$x = \pm \sqrt{\frac{- 1 - \sqrt{5}}{2}} = \pm \sqrt{\frac{1 + \sqrt{5}}{2}} i$