Solve left(x-2right)^2-4xleft(x-2right)=0

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

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=2$$

Steps Using Factoring By Grouping

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

$$x^{2}-4x+4-4x\left(x-2\right)=0$$

Use the distributive property to multiply -4x by x-2.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=4 ab=-3\times 4=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=6 b=-2$$

Rewrite -3x^{2}+4x+4 as \left(-3x^{2}+6x\right)+\left(-2x+4\right).

$$\left(-3x^{2}+6x\right)+\left(-2x+4\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(-x+2\right)+2\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(3x+2\right)$$

To find equation solutions, solve -x+2=0 and 3x+2=0.

$$x=2 x=-\frac{2}{3}$$

Steps Using the Quadratic Formula

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

$$x^{2}-4x+4-4x\left(x-2\right)=0$$

Use the distributive property to multiply -4x by x-2.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

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

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 4 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12\times 4}}{2\left(-3\right)}$$

Multiply 12 times 4.

$$x=\frac{-4±\sqrt{16+48}}{2\left(-3\right)}$$

Add 16 to 48.

$$x=\frac{-4±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-4±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-4±8}{-6}$$

Now solve the equation x=\frac{-4±8}{-6} when ± is plus. Add -4 to 8.

$$x=\frac{4}{-6}$$

Reduce the fraction \frac{4}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{3}$$

Now solve the equation x=\frac{-4±8}{-6} when ± is minus. Subtract 8 from -4.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=-\frac{2}{3} x=2$$

Steps for Completing the Square

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

$$x^{2}-4x+4-4x\left(x-2\right)=0$$

Use the distributive property to multiply -4x by x-2.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

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

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

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

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

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

Divide both sides by -3.

$$\frac{-3x^{2}+4x}{-3}=-\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{4}{-3}x=-\frac{4}{-3}$$

Divide 4 by -3.

$$x^{2}-\frac{4}{3}x=-\frac{4}{-3}$$

Divide -4 by -3.

$$x^{2}-\frac{4}{3}x=\frac{4}{3}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{4}{3}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{3}+\frac{4}{9}$$

Add \frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{2}{3}=\frac{4}{3} x-\frac{2}{3}=-\frac{4}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$x=2 x=-\frac{2}{3}$$