Negative Exponents in quadratic equation using U-substitute

Heres the equation:
x^-8 - 17x^-4 + 16 = 0

The book asks me to use U-Substitution, so here is my work:
u = x^-4 (u-substitution)
u^2 - 17u + 16 = 0
(u - 1) (u - 16) = 0
u = 1 u = 16
x^-4 = 1 x^-4 = 16

This is where I get stuck. I know that x = +/- 1 and x = +/- 1/2.
But the back of my book also says that x = +/- i and x = +/- 1/2i.
I don't understand where these imaginary numbers are coming from?

This equation has 8 solutions...4 real and 4 non real. Your solutions are correct.

\(\displaystyle x=i, \;\ x=\frac{1}{2}i, \;\ -i, \;\ \frac{-1}{2}i, \;\ 1, \;\ -1, \;\ \frac{1}{2}, \;\ \frac{-1}{2}\)

epitaph said:

Heres the equation:
x^-8 - 17x^-4 + 16 = 0

The book asks me to use U-Substitution, so here is my work:
u = x^-4 (u-substitution)
u^2 - 17u + 16 = 0
(u - 1) (u - 16) = 0
u = 1 u = 16
x^-4 = 1 x^-4 = 16

This is where I get stuck. I know that x = +/- 1 and x = +/- 1/2.
But the back of my book also says that x = +/- i and x = +/- 1/2i.
I don't understand where these imaginary numbers are coming from?

Click to expand...

\(\displaystyle \frac{1}{x^4} \, - \, 1 \, = \, 0\)

\(\displaystyle (\frac{1}{x^2} \, - \, 1)\cdot( \frac{1}{x^2} \, + \, 1) \, = \, 0\)

\(\displaystyle (\frac{1}{x^2} \, - \, 1)\cdot( \frac{1}{x^2} \, - \, i^2) \, = \, 0\)

\(\displaystyle (\frac{1}{x} \, - \, 1)\cdot( \frac{1}{x} \, + \, 1) \cdot(\frac{1}{x} \, - \, i)\cdot( \frac{1}{x} \, + \, i) \, = \, 0\)

This will give you four solutions. The other factor will give you the other four that galactus mentioned above.