# How do you solve for all solutions, including complex solutions, to x^4 + 4x^2 = 45?

May 2, 2016

$x = \pm \sqrt{5}$ or $x = \pm 3 i$

#### Explanation:

OK, let's complete the square to start, adding $4$ to both sides to get:

${x}^{4} + 4 {x}^{2} + 4 = 49$

That is:

${\left({x}^{2} + 2\right)}^{2} = {7}^{2}$

Hence:

${x}^{2} + 2 = \pm 7$

Subtract $2$ from both sides to get:

${x}^{2} = - 2 \pm 7$

That is:

${x}^{2} = 5$ or ${x}^{2} = - 9$

If ${x}^{2} = 5$ then $x = \pm \sqrt{5}$

If ${x}^{2} = - 9$ then $x = \pm 3 i$