Solve (x-1)^2=4x(x-1)

$$( x - 1 ) ^ { 2 } = 4 x ( x - 1 )$$

Solve for x

$$x=1$$

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1=4x\left(x-1\right)$$

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

$$x^{2}-2x+1=4x^{2}-4x$$

Subtract 4x^{2} from both sides.

$$x^{2}-2x+1-4x^{2}=-4x$$

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

$$-3x^{2}-2x+1=-4x$$

Add 4x to both sides.

$$-3x^{2}-2x+1+4x=0$$

Combine -2x and 4x to get 2x.

$$-3x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -3x^{2}+2x+1 as \left(-3x^{2}+3x\right)+\left(-x+1\right).

$$\left(-3x^{2}+3x\right)+\left(-x+1\right)$$

Factor out 3x in -3x^{2}+3x.

$$3x\left(-x+1\right)-x+1$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(3x+1\right)$$

To find equation solutions, solve -x+1=0 and 3x+1=0.

$$x=1 x=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1=4x\left(x-1\right)$$

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

$$x^{2}-2x+1=4x^{2}-4x$$

Subtract 4x^{2} from both sides.

$$x^{2}-2x+1-4x^{2}=-4x$$

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

$$-3x^{2}-2x+1=-4x$$

Add 4x to both sides.

$$-3x^{2}-2x+1+4x=0$$

Combine -2x and 4x to get 2x.

$$-3x^{2}+2x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -3.

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

Add 4 to 12.

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

Take the square root of 16.

$$x=\frac{-2±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±4}{-6}$$

Now solve the equation x=\frac{-2±4}{-6} when ± is plus. Add -2 to 4.

$$x=\frac{2}{-6}$$

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

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-2±4}{-6} when ± is minus. Subtract 4 from -2.

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

Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=-\frac{1}{3} x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1=4x\left(x-1\right)$$

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

$$x^{2}-2x+1=4x^{2}-4x$$

Subtract 4x^{2} from both sides.

$$x^{2}-2x+1-4x^{2}=-4x$$

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

$$-3x^{2}-2x+1=-4x$$

Add 4x to both sides.

$$-3x^{2}-2x+1+4x=0$$

Combine -2x and 4x to get 2x.

$$-3x^{2}+2x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -3.

$$\frac{-3x^{2}+2x}{-3}=-\frac{1}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{2}{-3}x=-\frac{1}{-3}$$

Divide 2 by -3.

$$x^{2}-\frac{2}{3}x=-\frac{1}{-3}$$

Divide -1 by -3.

$$x^{2}-\frac{2}{3}x=\frac{1}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$$

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{2}{3} x-\frac{1}{3}=-\frac{2}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=1 x=-\frac{1}{3}$$