# How do you find the real solutions of the polynomial x^4+2x^2=15?

##### 1 Answer
May 24, 2017

$x = \pm \sqrt{3}$

#### Explanation:

${x}^{4} + 2 {x}^{2} = 15$

Let $z = {x}^{2}$

$\therefore {z}^{2} + 2 z - 15 = 0$

$\left(z + 5\right) \left(z - 3\right) = 0$

$z = - 5$ or $+ 3$

Hence: $x = \pm \sqrt{- 5}$ or $x = \pm \sqrt{3}$

The question asks for real solutions only.

$\therefore x = \pm \sqrt{3}$