Solve (3x-1)^2=1

$$( 3 x - 1 ) ^ { 2 } = 1$$

Solve for x

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

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=1$$

Subtract 1 from both sides.

$$9x^{2}-6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$9x^{2}-6x=0$$

Factor out x.

$$x\left(9x-6\right)=0$$

To find equation solutions, solve x=0 and 9x-6=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=1$$

Subtract 1 from both sides.

$$9x^{2}-6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$9x^{2}-6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\times 9}$$

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±6}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±6}{18}$$

Now solve the equation x=\frac{6±6}{18} when ± is plus. Add 6 to 6.

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{6±6}{18} when ± is minus. Subtract 6 from 6.

$$x=\frac{0}{18}$$

Divide 0 by 18.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=1$$

Subtract 1 from both sides.

$$9x^{2}-6x=1-1$$

Subtract 1 from 1 to get 0.

$$9x^{2}-6x=0$$

Divide both sides by 9.

$$\frac{9x^{2}-6x}{9}=\frac{0}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{6}{9}\right)x=\frac{0}{9}$$

Reduce the fraction \frac{-6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=\frac{0}{9}$$

Divide 0 by 9.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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