Solve left(2x-1right)^2-1=0

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

Solve for x

$$x=1$$

$$x=0$$

Steps Using Factoring

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

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

Subtract 1 from 1 to get 0.

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

Factor out x.

$$x\left(4x-4\right)=0$$

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

$$x=0 x=1$$

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}.

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

Subtract 1 from 1 to get 0.

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

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

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±4}{8}$$

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

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

Divide 8 by 8.

$$x=1$$

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

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

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=1 x=0$$

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}.

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

Subtract 1 from 1 to get 0.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -4 by 4.

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

Divide 0 by 4.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=1 x=0$$