Algebra 2 Help......?

Let x^2 = y
y^2 - 8x - 9 = 0
(y-9)(y+1) = 0
Sub x^2 = y because remember we are trying to find x not y
(x^2-9)(x^2+1)=0
(x+3)(x-3)(x^2+1)=0
x=-3 or 3 or -i or i
so B

x^4 – 8x^2 – 9 = 0

(x-3)(x+3)(x^2 + 1) = 0

Then x = 3, -3, i or -i

Answer is b)

x^4 – 8x^2 – 9 = 0

(x² - 9)(x² + 1) = 0

x = ±3, ±i

x^4 - 8x^2 - 9 = 0
x^4 + x^2 - 9x^2 - 9 = 0
(x^4 + x^2) - (9x^2 + 9) = 0
x^2(x^2 + 1) - 9(x^2 + 1) = 0
(x^2 + 1)(x^2 - 9) = 0
[x^2 - (-1)](x^2 - 3^2) = 0
(x^2 - i^2)(x + 3)(x - 3) = 0
(x + i)(x - i)(x + 3)(x - 3) = 0

x + i = 0
x = -1

x - i = 0
x = i

x + 3 = 0
x = -3

x - 3 = 0
x = 3

∴ x = ±i, ±3
(answer b)

x^2's=[4+&-(16+9)^1/2]=9 & -1
x's=-3 , 3 , i & -i
Neighter one is compelete.
God bless you.

(x^2-9)(x^2+1)=0

x=-3, 3, and then I'm guessing i, -i though it have no idea what it means but that's the only thing that kinda makes sense. Because you need x^2 to equal -1 and that's basically impossible.

So I say B is the answer.