Solve 9(2x-1)^2-16(3x+1)^2=0

$$9 ( 2 x - 1 ) ^ { 2 } - 16 ( 3 x + 1 ) ^ { 2 } = 0$$

Solve for x

$$x=-\frac{1}{18}\approx -0.055555556$$

$$x = -\frac{7}{6} = -1\frac{1}{6} \approx -1.166666667$$

Steps Using Factoring By Grouping

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

$$9\left(4x^{2}-4x+1\right)-16\left(3x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$36x^{2}-36x+9-16\left(3x+1\right)^{2}=0$$

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

$$36x^{2}-36x+9-16\left(9x^{2}+6x+1\right)=0$$

Use the distributive property to multiply -16 by 9x^{2}+6x+1.

$$36x^{2}-36x+9-144x^{2}-96x-16=0$$

Combine 36x^{2} and -144x^{2} to get -108x^{2}.

$$-108x^{2}-36x+9-96x-16=0$$

Combine -36x and -96x to get -132x.

$$-108x^{2}-132x+9-16=0$$

Subtract 16 from 9 to get -7.

$$-108x^{2}-132x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -108x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=-132 ab=-108\left(-7\right)=756$$

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 756.

$$-1,-756 -2,-378 -3,-252 -4,-189 -6,-126 -7,-108 -9,-84 -12,-63 -14,-54 -18,-42 -21,-36 -27,-28$$

Calculate the sum for each pair.

$$-1-756=-757 -2-378=-380 -3-252=-255 -4-189=-193 -6-126=-132 -7-108=-115 -9-84=-93 -12-63=-75 -14-54=-68 -18-42=-60 -21-36=-57 -27-28=-55$$

The solution is the pair that gives sum -132.

$$a=-6 b=-126$$

Rewrite -108x^{2}-132x-7 as \left(-108x^{2}-6x\right)+\left(-126x-7\right).

$$\left(-108x^{2}-6x\right)+\left(-126x-7\right)$$

Factor out 6x in the first and 7 in the second group.

$$6x\left(-18x-1\right)+7\left(-18x-1\right)$$

Factor out common term -18x-1 by using distributive property.

$$\left(-18x-1\right)\left(6x+7\right)$$

To find equation solutions, solve -18x-1=0 and 6x+7=0.

$$x=-\frac{1}{18} x=-\frac{7}{6}$$

Steps Using the Quadratic Formula

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

$$9\left(4x^{2}-4x+1\right)-16\left(3x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$36x^{2}-36x+9-16\left(3x+1\right)^{2}=0$$

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

$$36x^{2}-36x+9-16\left(9x^{2}+6x+1\right)=0$$

Use the distributive property to multiply -16 by 9x^{2}+6x+1.

$$36x^{2}-36x+9-144x^{2}-96x-16=0$$

Combine 36x^{2} and -144x^{2} to get -108x^{2}.

$$-108x^{2}-36x+9-96x-16=0$$

Combine -36x and -96x to get -132x.

$$-108x^{2}-132x+9-16=0$$

Subtract 16 from 9 to get -7.

$$-108x^{2}-132x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -108 for a, -132 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-132\right)±\sqrt{\left(-132\right)^{2}-4\left(-108\right)\left(-7\right)}}{2\left(-108\right)}$$

Square -132.

$$x=\frac{-\left(-132\right)±\sqrt{17424-4\left(-108\right)\left(-7\right)}}{2\left(-108\right)}$$

Multiply -4 times -108.

$$x=\frac{-\left(-132\right)±\sqrt{17424+432\left(-7\right)}}{2\left(-108\right)}$$

Multiply 432 times -7.

$$x=\frac{-\left(-132\right)±\sqrt{17424-3024}}{2\left(-108\right)}$$

Add 17424 to -3024.

$$x=\frac{-\left(-132\right)±\sqrt{14400}}{2\left(-108\right)}$$

Take the square root of 14400.

$$x=\frac{-\left(-132\right)±120}{2\left(-108\right)}$$

The opposite of -132 is 132.

$$x=\frac{132±120}{2\left(-108\right)}$$

Multiply 2 times -108.

$$x=\frac{132±120}{-216}$$

Now solve the equation x=\frac{132±120}{-216} when ± is plus. Add 132 to 120.

$$x=\frac{252}{-216}$$

Reduce the fraction \frac{252}{-216} to lowest terms by extracting and canceling out 36.

$$x=-\frac{7}{6}$$

Now solve the equation x=\frac{132±120}{-216} when ± is minus. Subtract 120 from 132.

$$x=\frac{12}{-216}$$

Reduce the fraction \frac{12}{-216} to lowest terms by extracting and canceling out 12.

$$x=-\frac{1}{18}$$

The equation is now solved.

$$x=-\frac{7}{6} x=-\frac{1}{18}$$

Steps for Completing the Square

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

$$9\left(4x^{2}-4x+1\right)-16\left(3x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}-4x+1.

$$36x^{2}-36x+9-16\left(3x+1\right)^{2}=0$$

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

$$36x^{2}-36x+9-16\left(9x^{2}+6x+1\right)=0$$

Use the distributive property to multiply -16 by 9x^{2}+6x+1.

$$36x^{2}-36x+9-144x^{2}-96x-16=0$$

Combine 36x^{2} and -144x^{2} to get -108x^{2}.

$$-108x^{2}-36x+9-96x-16=0$$

Combine -36x and -96x to get -132x.

$$-108x^{2}-132x+9-16=0$$

Subtract 16 from 9 to get -7.

$$-108x^{2}-132x-7=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$-108x^{2}-132x=7$$

Divide both sides by -108.

$$\frac{-108x^{2}-132x}{-108}=\frac{7}{-108}$$

Dividing by -108 undoes the multiplication by -108.

$$x^{2}+\left(-\frac{132}{-108}\right)x=\frac{7}{-108}$$

Reduce the fraction \frac{-132}{-108} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{11}{9}x=\frac{7}{-108}$$

Divide 7 by -108.

$$x^{2}+\frac{11}{9}x=-\frac{7}{108}$$

Divide \frac{11}{9}, the coefficient of the x term, by 2 to get \frac{11}{18}. Then add the square of \frac{11}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{9}x+\left(\frac{11}{18}\right)^{2}=-\frac{7}{108}+\left(\frac{11}{18}\right)^{2}$$

Square \frac{11}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{9}x+\frac{121}{324}=-\frac{7}{108}+\frac{121}{324}$$

Add -\frac{7}{108} to \frac{121}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{9}x+\frac{121}{324}=\frac{25}{81}$$

Factor x^{2}+\frac{11}{9}x+\frac{121}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{18}\right)^{2}=\frac{25}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{18}\right)^{2}}=\sqrt{\frac{25}{81}}$$

Simplify.

$$x+\frac{11}{18}=\frac{5}{9} x+\frac{11}{18}=-\frac{5}{9}$$

Subtract \frac{11}{18} from both sides of the equation.

$$x=-\frac{1}{18} x=-\frac{7}{6}$$