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

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

Solve for x

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

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

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

$$8x^{2}-12x+4=-2x+1$$

Add 2x to both sides.

$$8x^{2}-12x+4+2x=1$$

Combine -12x and 2x to get -10x.

$$8x^{2}-10x+4=1$$

Subtract 1 from both sides.

$$8x^{2}-10x+4-1=0$$

Subtract 1 from 4 to get 3.

$$8x^{2}-10x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-10 ab=8\times 3=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite 8x^{2}-10x+3 as \left(8x^{2}-6x\right)+\left(-4x+3\right).

$$\left(8x^{2}-6x\right)+\left(-4x+3\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(4x-3\right)-\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(2x-1\right)$$

To find equation solutions, solve 4x-3=0 and 2x-1=0.

$$x=\frac{3}{4} x=\frac{1}{2}$$

Steps Using the Quadratic Formula

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

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

$$8x^{2}-12x+4=-2x+1$$

Add 2x to both sides.

$$8x^{2}-12x+4+2x=1$$

Combine -12x and 2x to get -10x.

$$8x^{2}-10x+4=1$$

Subtract 1 from both sides.

$$8x^{2}-10x+4-1=0$$

Subtract 1 from 4 to get 3.

$$8x^{2}-10x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 8\times 3}}{2\times 8}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 8\times 3}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-10\right)±\sqrt{100-32\times 3}}{2\times 8}$$

Multiply -32 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 8}$$

Add 100 to -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 8}$$

Take the square root of 4.

$$x=\frac{-\left(-10\right)±2}{2\times 8}$$

The opposite of -10 is 10.

$$x=\frac{10±2}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{10±2}{16}$$

Now solve the equation x=\frac{10±2}{16} when ± is plus. Add 10 to 2.

$$x=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{4}$$

Now solve the equation x=\frac{10±2}{16} when ± is minus. Subtract 2 from 10.

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

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

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

The equation is now solved.

$$x=\frac{3}{4} x=\frac{1}{2}$$

Steps for Completing the Square

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

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

$$8x^{2}-12x+4=-2x+1$$

Add 2x to both sides.

$$8x^{2}-12x+4+2x=1$$

Combine -12x and 2x to get -10x.

$$8x^{2}-10x+4=1$$

Subtract 4 from both sides.

$$8x^{2}-10x=1-4$$

Subtract 4 from 1 to get -3.

$$8x^{2}-10x=-3$$

Divide both sides by 8.

$$\frac{8x^{2}-10x}{8}=-\frac{3}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{10}{8}\right)x=-\frac{3}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{4}x=-\frac{3}{8}$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{3}{8}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{3}{8}+\frac{25}{64}$$

Add -\frac{3}{8} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{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}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{1}{8} x-\frac{5}{8}=-\frac{1}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=\frac{3}{4} x=\frac{1}{2}$$