Please help!!
Solve the equation and express each solution in a+bi form.
x^4-7x^2-8=0
a. x = -1, 1, 2 sqrt 2, or -2 sqrt 2
b. x = -i, i, 2 sqrt 2, or -2 sqrt 2
c. x = -1, 1, 2 sqrt 2i, or -2 sqrt 2i
d. x = -i, i, 2 sqrt 2i, or -2 sqrt 2i
Because this is a fourth degree that will be tricky to factor as it is, we will do a u substitution. Let [tex] x^2=u [/tex]. We can now rewrite that polynomial in terms of u: [tex] u^2-7u-8=0 [/tex]. Filling into the quadratic formula we have [tex] u=\frac{7+/-\sqrt{49-4(1)(-8)}}{2} [/tex]. Simplifying down [tex] u=\frac{7+/-\sqrt{81}}{2} [/tex] and [tex] u=\frac{7+9}{2} [/tex] or [tex] u=\frac{7-9}{2} [/tex]. That means that u = 8 or u = -1. But don't forget that we let [tex] u=x^2 [/tex], so we have to put x-squared back in for u. That gives us [tex] x^2=8 [/tex] which simplifies down to [tex] +/-2\sqrt{2} [/tex]. That also gives us [tex] x^2=-1 [/tex]. When we take the square root of -1, we have to use the fact that -1 = i^2, so we sub that in to get [tex] x=+/-i [/tex]. All in all, your solutions are as follows: [tex] 2\sqrt{2},-2\sqrt{2},i,-i [/tex] which is choice b.
