Solve left(2x-2right)^2-4x(x-1)>0

$${ \left(2x-2 \right) }^{ 2 } -4x(x-1) > 0$$

Solve for x

$$x<1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-2\right)^{2}.

$$4x^{2}-8x+4-4x\left(x-1\right)>0$$

Use the distributive property to multiply -4x by x-1.

$$4x^{2}-8x+4-4x^{2}+4x>0$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-8x+4+4x>0$$

Combine -8x and 4x to get -4x.

$$-4x+4>0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-4x>-4$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<\frac{-4}{-4}$$

Divide -4 by -4 to get 1.

$$x<1$$