What are the solutions to the equation (x^2-2x)^2-(x^2-2x)=2 ?

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

Find the zeros of (x^2-2x)^2-(x^2-2x)=2

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

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

x = 1, x = 1 + 3, x = 1 - 3

Solution steps

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

Expand

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

x^{4} - 4 x^{3} + 3 x^{2} + 2 x = 2

Move

2

to the left side

x^{4} - 4 x^{3} + 3 x^{2} + 2 x - 2 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x^{2} - 2 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

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

The solutions are

x = 1, x = 1 + 3, x = 1 - 3

x^{3} + 6 x - 20 = 0

x^{3} - 9 x^{2} - 81 x - 729 = 0

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

2 x^{4} - 21 x^{2} + 45 = 0

2 x^{3} = x^{2}

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