Solve (3x-1)^2=x^2

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

Solve for x

$$x=\frac{1}{4}=0.25$$

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

Steps Using Factoring By Grouping

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

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-6x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-6 ab=8\times 1=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite 8x^{2}-6x+1 as \left(8x^{2}-4x\right)+\left(-2x+1\right).

$$\left(8x^{2}-4x\right)+\left(-2x+1\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(2x-1\right)-\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(4x-1\right)$$

To find equation solutions, solve 2x-1=0 and 4x-1=0.

$$x=\frac{1}{2} x=\frac{1}{4}$$

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}.

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-6x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -6 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32}}{2\times 8}$$

Add 36 to -32.

$$x=\frac{-\left(-6\right)±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

$$x=\frac{-\left(-6\right)±2}{2\times 8}$$

The opposite of -6 is 6.

$$x=\frac{6±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±2}{16}$$

Now solve the equation x=\frac{6±2}{16} when ± is plus. Add 6 to 2.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{6±2}{16} when ± is minus. Subtract 2 from 6.

$$x=\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{1}{2} x=\frac{1}{4}$$

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}.

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

Subtract x^{2} from both sides.

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

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-6x+1=0$$

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

$$8x^{2}-6x=-1$$

Divide both sides by 8.

$$\frac{8x^{2}-6x}{8}=-\frac{1}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{6}{8}\right)x=-\frac{1}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{4}x=-\frac{1}{8}$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-\frac{1}{8}+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{1}{8}+\frac{9}{64}$$

Add -\frac{1}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{1}{8} x-\frac{3}{8}=-\frac{1}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{1}{2} x=\frac{1}{4}$$