Help with a polynomial question? Solve?

Question

x^-4 + 8x^-2 +15 = 0

Answer 1

The given equation is quadratic in form.
Let u = x^(-2). Then u^2 = x^(-4).

The equation becomes
u^2 + 8u + 15 = 0.
(u + 5)(u + 3) = 0
u + 5 = 0 ||||| u + 3 =0
u = -5 ||||| u = -3

Then we return it to the orig variable: u = x^(-2)
u = x^(-2) = -5
x^2 = (-5)^(-1) *mult both sides by -1
x^2 = -1/5
x = sqrt (-1/5)
x = (+or -) i sqrt(1/5)

u = x^(-2) = -3
x^2 = (-3)^(-1)
x^2 = -1/3
x = sqrt (-1/3)
x = (+or -) i sqrt(1/3)

There are four solutions:
x = (+or -) i sqrt(1/5)
x = (+or -) i sqrt(1/3)

Answer 2

x^-4 + 8x^-2 +15 = 0

First multiply both sides by x^4:
1 + 8x^2 + 15x^4 = 0

Rearrange terms in descending order:
15x^4 + 8x^2 + 1 = 0

Let u = x^2
15u^2 + 8u + 1 = 0

Factor:
(5u + 1)(3u + 1) = 0
5u + 1 = 0 and 3u + 1 = 0

5x^2 + 1 = 0
x^2 = -1/5
x = ±sqrt(-1/5) = ±i*sqrt(1/5)
and
3x^2 + 1 = 0
x^2 = -1/3
x = ±sqrt(-1/3) = ±i*sqrt(1/3)

No real solution, but four complex solutions.

Answer 3

x^-4 + 8x^-2 +15 = 0

Let y = x^-2

y^2 + 8y + 15 = 0
(y + 3)(y + 5) = 0

y = -3, y = -5

x^-2 = -3
x^2 = - 1/3
x = +- sqrt(-1/3)

x^-2 = -5
x = +- sqrt(-1/5)

Answer 4

multiplying the equation by x^4 gives
1+ 8x^2 +15x^4 = 0
let y=x^2
so, 1+8y+15y^2=0
i.e. 15y^2 + 5y+ 3y + 1=0
i.e. (5y+1)(3y+1)= 0
so
y= -1/5 or y= -1/3
so
x^2=-1/5 or x^2= -1/3

so, x= +iota*(1/5)^-2
or, x= -iota*(1/5)^-2
or, x= +iota*(1/3)^-2
or, x= -iota*(1/3)^-2

this equation has 4 solutions, all imaginary.

Answer 5

(x^-2 + 5)(x^-2 + 3) = 0
x^-2 = -5, x^2 = -1/5, x = ±i√5
x^-2 = -3, x^2 = -1/3, x = ±i√3

Answer 6

9x = 7
x=7/9
then you have to do the sqaure root of what ever that power is supposed to be