What are the solutions to the equation (x^2-x)^2-8(x^2-x)+12=0 ?

The solutions to the equation (x^2-x)^2-8(x^2-x)+12=0 are x=-1,x=3,x=-2,x=2

Find the zeros of (x^2-x)^2-8(x^2-x)+12=0

The zeros of (x^2-x)^2-8(x^2-x)+12=0 are x=-1,x=3,x=-2,x=2

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(x^{2} - x)^{2} - 8 (x^{2} - x) + 12 = 0

x = - 1, x = 3, x = - 2, x = 2

Solution steps

(x^{2} - x)^{2} - 8 (x^{2} - x) + 12 = 0

Expand

(x^{2} - x)^{2} - 8 (x^{2} - x) + 12 : x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12

x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12 = 0

Factor

x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12 : (x + 1) (x - 3) (x + 2) (x - 2)

(x + 1) (x - 3) (x + 2) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 3 = 0 or x + 2 = 0 or x - 2 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 3 = 0 : x = 3

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = - 1, x = 3, x = - 2, x = 2

y = (y^{2})^{2}

x (x + 2) (x + 4) = 4

n^{4} - 5 n^{2} = - 4

x^{3} + 7 x^{2} + 25 x + 175 = 0

x (x - 8) (x - 4)^{2} = 225

What are the solutions to the equation (x^2-x)^2-8(x^2-x)+12=0 ?
The solutions to the equation (x^2-x)^2-8(x^2-x)+12=0 are x=-1,x=3,x=-2,x=2
Find the zeros of (x^2-x)^2-8(x^2-x)+12=0
The zeros of (x^2-x)^2-8(x^2-x)+12=0 are x=-1,x=3,x=-2,x=2