Solve (1)(3x+1)^2=9(2x+3)^2

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

Solve for x

$$x = -\frac{10}{9} = -1\frac{1}{9} \approx -1.111111111$$

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

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}.

$$1\left(9x^{2}+6x+1\right)=9\left(2x+3\right)^{2}$$

Use the distributive property to multiply 1 by 9x^{2}+6x+1.

$$9x^{2}+6x+1=9\left(2x+3\right)^{2}$$

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

$$9x^{2}+6x+1=9\left(4x^{2}+12x+9\right)$$

Use the distributive property to multiply 9 by 4x^{2}+12x+9.

$$9x^{2}+6x+1=36x^{2}+108x+81$$

Subtract 36x^{2} from both sides.

$$9x^{2}+6x+1-36x^{2}=108x+81$$

Combine 9x^{2} and -36x^{2} to get -27x^{2}.

$$-27x^{2}+6x+1=108x+81$$

Subtract 108x from both sides.

$$-27x^{2}+6x+1-108x=81$$

Combine 6x and -108x to get -102x.

$$-27x^{2}-102x+1=81$$

Subtract 81 from both sides.

$$-27x^{2}-102x+1-81=0$$

Subtract 81 from 1 to get -80.

$$-27x^{2}-102x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -27 for a, -102 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-102\right)±\sqrt{\left(-102\right)^{2}-4\left(-27\right)\left(-80\right)}}{2\left(-27\right)}$$

Square -102.

$$x=\frac{-\left(-102\right)±\sqrt{10404-4\left(-27\right)\left(-80\right)}}{2\left(-27\right)}$$

Multiply -4 times -27.

$$x=\frac{-\left(-102\right)±\sqrt{10404+108\left(-80\right)}}{2\left(-27\right)}$$

Multiply 108 times -80.

$$x=\frac{-\left(-102\right)±\sqrt{10404-8640}}{2\left(-27\right)}$$

Add 10404 to -8640.

$$x=\frac{-\left(-102\right)±\sqrt{1764}}{2\left(-27\right)}$$

Take the square root of 1764.

$$x=\frac{-\left(-102\right)±42}{2\left(-27\right)}$$

The opposite of -102 is 102.

$$x=\frac{102±42}{2\left(-27\right)}$$

Multiply 2 times -27.

$$x=\frac{102±42}{-54}$$

Now solve the equation x=\frac{102±42}{-54} when ± is plus. Add 102 to 42.

$$x=\frac{144}{-54}$$

Reduce the fraction \frac{144}{-54} to lowest terms by extracting and canceling out 18.

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

Now solve the equation x=\frac{102±42}{-54} when ± is minus. Subtract 42 from 102.

$$x=\frac{60}{-54}$$

Reduce the fraction \frac{60}{-54} to lowest terms by extracting and canceling out 6.

$$x=-\frac{10}{9}$$

The equation is now solved.

$$x=-\frac{8}{3} x=-\frac{10}{9}$$

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}.

$$1\left(9x^{2}+6x+1\right)=9\left(2x+3\right)^{2}$$

Use the distributive property to multiply 1 by 9x^{2}+6x+1.

$$9x^{2}+6x+1=9\left(2x+3\right)^{2}$$

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

$$9x^{2}+6x+1=9\left(4x^{2}+12x+9\right)$$

Use the distributive property to multiply 9 by 4x^{2}+12x+9.

$$9x^{2}+6x+1=36x^{2}+108x+81$$

Subtract 36x^{2} from both sides.

$$9x^{2}+6x+1-36x^{2}=108x+81$$

Combine 9x^{2} and -36x^{2} to get -27x^{2}.

$$-27x^{2}+6x+1=108x+81$$

Subtract 108x from both sides.

$$-27x^{2}+6x+1-108x=81$$

Combine 6x and -108x to get -102x.

$$-27x^{2}-102x+1=81$$

Subtract 1 from both sides.

$$-27x^{2}-102x=81-1$$

Subtract 1 from 81 to get 80.

$$-27x^{2}-102x=80$$

Divide both sides by -27.

$$\frac{-27x^{2}-102x}{-27}=\frac{80}{-27}$$

Dividing by -27 undoes the multiplication by -27.

$$x^{2}+\left(-\frac{102}{-27}\right)x=\frac{80}{-27}$$

Reduce the fraction \frac{-102}{-27} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{34}{9}x=\frac{80}{-27}$$

Divide 80 by -27.

$$x^{2}+\frac{34}{9}x=-\frac{80}{27}$$

Divide \frac{34}{9}, the coefficient of the x term, by 2 to get \frac{17}{9}. Then add the square of \frac{17}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{34}{9}x+\left(\frac{17}{9}\right)^{2}=-\frac{80}{27}+\left(\frac{17}{9}\right)^{2}$$

Square \frac{17}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{34}{9}x+\frac{289}{81}=-\frac{80}{27}+\frac{289}{81}$$

Add -\frac{80}{27} to \frac{289}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{34}{9}x+\frac{289}{81}=\frac{49}{81}$$

Factor x^{2}+\frac{34}{9}x+\frac{289}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{17}{9}\right)^{2}=\frac{49}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{9}\right)^{2}}=\sqrt{\frac{49}{81}}$$

Simplify.

$$x+\frac{17}{9}=\frac{7}{9} x+\frac{17}{9}=-\frac{7}{9}$$

Subtract \frac{17}{9} from both sides of the equation.

$$x=-\frac{10}{9} x=-\frac{8}{3}$$