What are the solutions to the equation (x^2+3x)^2-2(x^2+3x)-8=0 ?

The solutions to the equation (x^2+3x)^2-2(x^2+3x)-8=0 are x=-1,x=1,x=-2,x=-4

Find the zeros of (x^2+3x)^2-2(x^2+3x)-8=0

The zeros of (x^2+3x)^2-2(x^2+3x)-8=0 are x=-1,x=1,x=-2,x=-4

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(x^{2} + 3 x)^{2} - 2 (x^{2} + 3 x) - 8 = 0

x = - 1, x = 1, x = - 2, x = - 4

Solution steps

(x^{2} + 3 x)^{2} - 2 (x^{2} + 3 x) - 8 = 0

Expand

(x^{2} + 3 x)^{2} - 2 (x^{2} + 3 x) - 8 : x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 = 0

Factor

x^{4} + 6 x^{3} + 7 x^{2} - 6 x - 8 : (x + 1) (x - 1) (x + 2) (x + 4)

(x + 1) (x - 1) (x + 2) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 1 = 0 or x + 2 = 0 or x + 4 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x + 4 = 0 : x = - 4

The solutions are

x = - 1, x = 1, x = - 2, x = - 4

216 x = x^{4}

g (n) = n^{3} + 4 n, g (4)

4 x^{4} = 10

1 - 3 x + 3 x^{2} - x^{3} = 0

x^{3} = - x + 1

What are the solutions to the equation (x^2+3x)^2-2(x^2+3x)-8=0 ?
The solutions to the equation (x^2+3x)^2-2(x^2+3x)-8=0 are x=-1,x=1,x=-2,x=-4
Find the zeros of (x^2+3x)^2-2(x^2+3x)-8=0
The zeros of (x^2+3x)^2-2(x^2+3x)-8=0 are x=-1,x=1,x=-2,x=-4