Solve (3x+1)^2=9

$$( 3 x + 1 ) ^ { 2 } = 9$$

Solve for x

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

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

Steps Using Factoring By Grouping

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

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$9x^{2}+6x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=9\left(-8\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=-6 b=12$$

Rewrite 9x^{2}+6x-8 as \left(9x^{2}-6x\right)+\left(12x-8\right).

$$\left(9x^{2}-6x\right)+\left(12x-8\right)$$

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

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

Factor out common term 3x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$9x^{2}+6x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-8\right)}}{2\times 9}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-8\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-6±\sqrt{36-36\left(-8\right)}}{2\times 9}$$

Multiply -36 times -8.

$$x=\frac{-6±\sqrt{36+288}}{2\times 9}$$

Add 36 to 288.

$$x=\frac{-6±\sqrt{324}}{2\times 9}$$

Take the square root of 324.

$$x=\frac{-6±18}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±18}{18}$$

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

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

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±18}{18} when ± is minus. Subtract 18 from -6.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Subtract 1 from both sides.

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

Subtract 1 from 9 to get 8.

$$9x^{2}+6x=8$$

Divide both sides by 9.

$$\frac{9x^{2}+6x}{9}=\frac{8}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{6}{9}x=\frac{8}{9}$$

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

$$x^{2}+\frac{2}{3}x=\frac{8}{9}$$

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{8}{9}+\left(\frac{1}{3}\right)^{2}$$

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

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{3} from both sides of the equation.

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