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

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

Solve for x

$$x=0$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring

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

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

Subtract 2x from both sides.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Factor out x.

$$x\left(4x+2\right)=0$$

To find equation solutions, solve x=0 and 4x+2=0.

$$x=0 x=-\frac{1}{2}$$

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}.

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

Subtract 2x from both sides.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 4}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2}{8}$$

Now solve the equation x=\frac{-2±2}{8} when ± is plus. Add -2 to 2.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

Now solve the equation x=\frac{-2±2}{8} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=0 x=-\frac{1}{2}$$

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}.

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

Subtract 2x from both sides.

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Divide both sides by 4.

$$\frac{4x^{2}+2x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{2}{4}x=\frac{0}{4}$$

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

$$x^{2}+\frac{1}{2}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}+\frac{1}{2}x=0$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{1}{4} x+\frac{1}{4}=-\frac{1}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=0 x=-\frac{1}{2}$$