What are the solutions to the equation (x^2+4)2-11(x^2+4)+24=0 ?

The solutions to the equation (x^2+4)2-11(x^2+4)+24=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i

Find the zeros of (x^2+4)2-11(x^2+4)+24=0

The zeros of (x^2+4)2-11(x^2+4)+24=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i

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(x^{2} + 4) 2 - 11 (x^{2} + 4) + 24 = 0

x = \frac{2}{3} i, x = - \frac{2}{3} i

Solution steps

(x^{2} + 4) \cdot 2 - 11 (x^{2} + 4) + 24 = 0

Move

24

to the right side

(x^{2} + 4) \cdot 2 - 11 (x^{2} + 4) = - 24

Add similar elements:

2 (x^{2} + 4) - 11 (x^{2} + 4) = - 9 (x^{2} + 4)

- 9 (x^{2} + 4) = - 24

Divide both sides by

- 9

x^{2} + 4 = \frac{8}{3}

Move

4

to the right side

x^{2} = - \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - \frac{4}{3}, x = - - \frac{4}{3}

- \frac{4}{3} = \frac{2}{3} i

- - \frac{4}{3} = - \frac{2}{3} i

x = \frac{2}{3} i, x = - \frac{2}{3} i

t^{2} + 12 t + 36 = 0

4 p^{2} - 12 p + 3 = 0

25 y^{2} = 36 y

5 x^{2} + 7 x = 2

t, s = - 16 t^{2} + 64 t

What are the solutions to the equation (x^2+4)2-11(x^2+4)+24=0 ?
The solutions to the equation (x^2+4)2-11(x^2+4)+24=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i
Find the zeros of (x^2+4)2-11(x^2+4)+24=0
The zeros of (x^2+4)2-11(x^2+4)+24=0 are x= 2/(sqrt(3))i,x=-2/(sqrt(3))i