Question

Find one value of x that is a solution to the equation:
(x^2+4)^2 – 11(x^2+4) + 24 = 0

Answer 1

one value of x = 2

Explanation:

`(x^2 + 4 )^2 - 11(x^2 + 4 ) + 24 = 0\\\\x^4 + 16 + 8x^2 - 11x^2 -44 + 24 = 0\\\\x^4 - 3x^2 -4 = 0\\\\` ------ ( 1 )

`Let \ x^2 \ = \ u`

( 1 ) =>
`u^2 - 3u - 4 = 0`

`u^2 -4u + u - 4 = 0\\\\u(u - 4) + 1 (u - 4) = 0\\\\(u + 1) (u - 4) = 0\\\\u = -1 \ , \ u = 4`

`=> x^2 = - 1 \ and \ x^2 = 4`

`x = \sqrt {-1} = i`

`x = √(4) = \pm 2`

Answer 2

x = ±2

x = ±i

Explanation:

(x^2+4)^2 – 11(x^2+4) + 24 = 0

Let m = x^2 +4

(m)^2 – 11(m) + 24 = 0

Solving this quadratic

What two numbers multiply to 24 and add to -11

-8*-3 =24

-8-3 = -11

(m-8)(m-3) =0

m = 8 m=3

Now substitute back

x^2 +4 = 8 x^2 +4 = 3

x^2 +4-4 = 8-4 x^2+4-4 = 3-4

x^2 = 4 x^2 = -1

Taking the square root

sqrt(x^2) = sqrt(4) sqrt(x^2) = sqrt(-1)

x = ±2 x = ±i