Question 210917
x^-8-17x^-4+16=0
.
Let t = x^-4
now, substitute the above into the original:
t^2 - 17t + 16 = 0
(t-16)(t-1) = 0
t = {1, 16}
.
To find x, use the fact that t=x^-4
1 = x^-4
1 = 1/x^4
x = 1
.
16 = x^-4
16 = 1/x^4
x^4 = 1/16
x = 1/2
.
x = +-{1, 1/2}