Solve (3x-2)^2=36

$$( 3 x - 2 ) ^ { 2 } = 36$$

Solve for x

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

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}.

$$9x^{2}-12x+4=36$$

Subtract 36 from both sides.

$$9x^{2}-12x+4-36=0$$

Subtract 36 from 4 to get -32.

$$9x^{2}-12x-32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=9\left(-32\right)=-288$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -288.

$$1,-288 2,-144 3,-96 4,-72 6,-48 8,-36 9,-32 12,-24 16,-18$$

Calculate the sum for each pair.

$$1-288=-287 2-144=-142 3-96=-93 4-72=-68 6-48=-42 8-36=-28 9-32=-23 12-24=-12 16-18=-2$$

The solution is the pair that gives sum -12.

$$a=-24 b=12$$

Rewrite 9x^{2}-12x-32 as \left(9x^{2}-24x\right)+\left(12x-32\right).

$$\left(9x^{2}-24x\right)+\left(12x-32\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(3x-8\right)+4\left(3x-8\right)$$

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(3x+4\right)$$

To find equation solutions, solve 3x-8=0 and 3x+4=0.

$$x=\frac{8}{3} x=-\frac{4}{3}$$

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}.

$$9x^{2}-12x+4=36$$

Subtract 36 from both sides.

$$9x^{2}-12x+4-36=0$$

Subtract 36 from 4 to get -32.

$$9x^{2}-12x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-32\right)}}{2\times 9}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\left(-32\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\left(-32\right)}}{2\times 9}$$

Multiply -36 times -32.

$$x=\frac{-\left(-12\right)±\sqrt{144+1152}}{2\times 9}$$

Add 144 to 1152.

$$x=\frac{-\left(-12\right)±\sqrt{1296}}{2\times 9}$$

Take the square root of 1296.

$$x=\frac{-\left(-12\right)±36}{2\times 9}$$

The opposite of -12 is 12.

$$x=\frac{12±36}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{12±36}{18}$$

Now solve the equation x=\frac{12±36}{18} when ± is plus. Add 12 to 36.

$$x=\frac{48}{18}$$

Reduce the fraction \frac{48}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{12±36}{18} when ± is minus. Subtract 36 from 12.

$$x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

$$x=\frac{8}{3} x=-\frac{4}{3}$$

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}.

$$9x^{2}-12x+4=36$$

Subtract 4 from both sides.

$$9x^{2}-12x=36-4$$

Subtract 4 from 36 to get 32.

$$9x^{2}-12x=32$$

Divide both sides by 9.

$$\frac{9x^{2}-12x}{9}=\frac{32}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{12}{9}\right)x=\frac{32}{9}$$

Reduce the fraction \frac{-12}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{4}{3}x=\frac{32}{9}$$

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

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{32}{9}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{32+4}{9}$$

Add \frac{32}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{8}{3} x=-\frac{4}{3}$$