Question

Find the real or imaginary solutions by factoring
x^4-8x^2=-16

Answer 1

x = +/- 2

Explanation:

`x^(4) - 8x^(2) = -16` Rewrite in form P(x) = 0 by adding 16 to both sides.

`x^(4) - 8x^(2) + 16 = 0` Write in terms of
`x^(2)`.

`(x^(2) )^(2) -8(x^(2) ) + 16 = 0` Let a =
`x^(2)`.

`a^(2) - 8a + 16 = 0` Factor the expression.

`(a - 4)(a - 4) = 0` Replace a with
`x^(2)`.

`(x^(2) - 4) (x^(2) - 4) = 0` Factor
`x^(2) - 4` as a difference of squares.

`(x -2)(x -2) (x -2)(x -2) = 0`

`x -2 = 0, x-2 = 0` Since it is in duplicity,

x = 2 or -2. or +/- 2.

Answer 2

x^4 - 8x^2 + 16 = 0
(x^2 - 4)(x^2 - 4) = 0

x^2 - 4 = 0
x= +2 , -2

Solution is x = 2 (duplicity 2) and x = -2 (duplicity 2)