Question

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

Answer 1

`x=+-2`, each with multiplicity `2`

Explanation:

Given:

`x^4-8x^2=-16`

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

`0 = x^4-8x^2+16 = (x^2-4)^2 = (x-2)^2(x+2)^2`

So the solutions of the equation are `x = +-2`, each with multiplicity `2`.