# How do you find the real and imaginary solutions of x^4-8x^2=-16?

Nov 28, 2016

$x = \pm 2$, each with multiplicity $2$

#### Explanation:

Given:

${x}^{4} - 8 {x}^{2} = - 16$

Add $16$ to both sides of the equation and transpose to get:

$0 = {x}^{4} - 8 {x}^{2} + 16 = {\left({x}^{2} - 4\right)}^{2} = {\left(x - 2\right)}^{2} {\left(x + 2\right)}^{2}$

So the solutions of the equation are $x = \pm 2$, each with multiplicity $2$.