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

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

Solve for x

$$x=-2$$

$$x=0$$

Steps Using Factoring

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

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$3x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$3x^{2}+6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$3x^{2}+6x=0$$

Factor out x.

$$x\left(3x+6\right)=0$$

To find equation solutions, solve x=0 and 3x+6=0.

$$x=0 x=-2$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$3x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$3x^{2}+6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$3x^{2}+6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 3}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-6±6}{6}$$

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

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

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

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

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

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

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

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

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

Subtract x^{2} from both sides.

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

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

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

Add 2x to both sides.

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

Combine 4x and 2x to get 6x.

$$3x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$3x^{2}+6x=1-1$$

Subtract 1 from 1 to get 0.

$$3x^{2}+6x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+6x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{6}{3}x=\frac{0}{3}$$

Divide 6 by 3.

$$x^{2}+2x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+2x=0$$

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

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

Square 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$