Can anyone help with a problem with negative exponents and the quadratic formula?

Question

144 x^-4 - 40 x^-2 + 1 =0

There's supposed to be 4 different answers (+ or -)

Man, I've been getting so upset trying to work this problem out. Thank you all for explaining how to do it. I will make a note on how to do problems like this in the future.

Answer 1

144 x^-4 - 40 x^-2 + 1 =0

u = x^-2

144 u^2 - 40u + 1 =0
u=1/4 , u =1/36

x^-2 = 1/4
x=±2
x^-2 = 1/32
x=±6

Answer 2

for the negatives just take the reciprocal of the number, then
use substitution
let z=x^2
(1/144z^2)-(1/40z)+1
the quadratic formula should help from this point on, or if you have a graphing calculator then use that and find the roots.

Answer 3

By the first fundamental theorem of algebra, four solutions exist. They may not be unique, but they exist in the domain of Complex numbers.

Don't be intimidated by the negative exponents. The mechanics of solving the problem are not affected by the negative exponents.

Answer 4

144x^(-4) - 40 x^(-2) + 1 = 0
[36x^(-2) - 1] [4x^(-2) - 1] = 0
[6x^(-1) + 1] [6x^(-1) - 1] [2x^(-1) + 1] [2x^(-1) - 1] = 0
(6/x + 1)(6/x - 1)(2/x + 1)(2/x - 1) = 0

x = ±6, ±2