What are the solutions to the equation (x^2-2x)^2-(x^2-2x)=6 ?

The solutions to the equation (x^2-2x)^2-(x^2-2x)=6 are x=-1,x=3,x=1+i,x=1-i

Find the zeros of (x^2-2x)^2-(x^2-2x)=6

The zeros of (x^2-2x)^2-(x^2-2x)=6 are x=-1,x=3,x=1+i,x=1-i

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(x^{2} - 2 x)^{2} - (x^{2} - 2 x) = 6

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

Solution steps

(x^{2} - 2 x)^{2} - (x^{2} - 2 x) = 6

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 = 6

Move

6

to the left side

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

Factor

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

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

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

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

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

What are the solutions to the equation (x^2-2x)^2-(x^2-2x)=6 ?
The solutions to the equation (x^2-2x)^2-(x^2-2x)=6 are x=-1,x=3,x=1+i,x=1-i
Find the zeros of (x^2-2x)^2-(x^2-2x)=6
The zeros of (x^2-2x)^2-(x^2-2x)=6 are x=-1,x=3,x=1+i,x=1-i