What are the solutions to the equation (x^2+3x)^2=4(x^2+3x) ?

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

Find the zeros of (x^2+3x)^2=4(x^2+3x)

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

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(x^{2} + 3 x)^{2} = 4 (x^{2} + 3 x)

x = 0, x = 1, x = - 3, x = - 4

Solution steps

(x^{2} + 3 x)^{2} = 4 (x^{2} + 3 x)

Expand

(x^{2} + 3 x)^{2} : x^{4} + 6 x^{3} + 9 x^{2}

Expand

4 (x^{2} + 3 x) : 4 x^{2} + 12 x

x^{4} + 6 x^{3} + 9 x^{2} = 4 x^{2} + 12 x

Move

12 x

to the left side

x^{4} + 6 x^{3} + 9 x^{2} - 12 x = 4 x^{2}

Move

4 x^{2}

to the left side

x^{4} + 6 x^{3} + 5 x^{2} - 12 x = 0

Factor

x^{4} + 6 x^{3} + 5 x^{2} - 12 x : x (x - 1) (x + 3) (x + 4)

x (x - 1) (x + 3) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x + 3 = 0 or x + 4 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 3 = 0 : x = - 3

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 0, x = 1, x = - 3, x = - 4

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

x^{2} (x - 2)^{2} (x - 1)^{2} = 0

x^{3} = 9000

2 = y^{3}

y^{4} + 27 y^{2} - 324 = 0

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