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

1 Answer
Nov 28, 2016

Answer:

#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#.