Solve 9left(3x+1right)^2-64=0

$$9 { \left(3x+1 \right) }^{ 2 } -64=0$$

Solve for x

$$x=\frac{5}{9}\approx 0.555555556$$

$$x = -\frac{11}{9} = -1\frac{2}{9} \approx -1.222222222$$

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}.

$$9\left(9x^{2}+6x+1\right)-64=0$$

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

$$81x^{2}+54x+9-64=0$$

Subtract 64 from 9 to get -55.

$$81x^{2}+54x-55=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 81x^{2}+ax+bx-55. To find a and b, set up a system to be solved.

$$a+b=54 ab=81\left(-55\right)=-4455$$

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 -4455.

$$-1,4455 -3,1485 -5,891 -9,495 -11,405 -15,297 -27,165 -33,135 -45,99 -55,81$$

Calculate the sum for each pair.

$$-1+4455=4454 -3+1485=1482 -5+891=886 -9+495=486 -11+405=394 -15+297=282 -27+165=138 -33+135=102 -45+99=54 -55+81=26$$

The solution is the pair that gives sum 54.

$$a=-45 b=99$$

Rewrite 81x^{2}+54x-55 as \left(81x^{2}-45x\right)+\left(99x-55\right).

$$\left(81x^{2}-45x\right)+\left(99x-55\right)$$

Factor out 9x in the first and 11 in the second group.

$$9x\left(9x-5\right)+11\left(9x-5\right)$$

Factor out common term 9x-5 by using distributive property.

$$\left(9x-5\right)\left(9x+11\right)$$

To find equation solutions, solve 9x-5=0 and 9x+11=0.

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

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}.

$$9\left(9x^{2}+6x+1\right)-64=0$$

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

$$81x^{2}+54x+9-64=0$$

Subtract 64 from 9 to get -55.

$$81x^{2}+54x-55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 81 for a, 54 for b, and -55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}-4\times 81\left(-55\right)}}{2\times 81}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\times 81\left(-55\right)}}{2\times 81}$$

Multiply -4 times 81.

$$x=\frac{-54±\sqrt{2916-324\left(-55\right)}}{2\times 81}$$

Multiply -324 times -55.

$$x=\frac{-54±\sqrt{2916+17820}}{2\times 81}$$

Add 2916 to 17820.

$$x=\frac{-54±\sqrt{20736}}{2\times 81}$$

Take the square root of 20736.

$$x=\frac{-54±144}{2\times 81}$$

Multiply 2 times 81.

$$x=\frac{-54±144}{162}$$

Now solve the equation x=\frac{-54±144}{162} when ± is plus. Add -54 to 144.

$$x=\frac{90}{162}$$

Reduce the fraction \frac{90}{162} to lowest terms by extracting and canceling out 18.

$$x=\frac{5}{9}$$

Now solve the equation x=\frac{-54±144}{162} when ± is minus. Subtract 144 from -54.

$$x=-\frac{198}{162}$$

Reduce the fraction \frac{-198}{162} to lowest terms by extracting and canceling out 18.

$$x=-\frac{11}{9}$$

The equation is now solved.

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

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}.

$$9\left(9x^{2}+6x+1\right)-64=0$$

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

$$81x^{2}+54x+9-64=0$$

Subtract 64 from 9 to get -55.

$$81x^{2}+54x-55=0$$

Add 55 to both sides. Anything plus zero gives itself.

$$81x^{2}+54x=55$$

Divide both sides by 81.

$$\frac{81x^{2}+54x}{81}=\frac{55}{81}$$

Dividing by 81 undoes the multiplication by 81.

$$x^{2}+\frac{54}{81}x=\frac{55}{81}$$

Reduce the fraction \frac{54}{81} to lowest terms by extracting and canceling out 27.

$$x^{2}+\frac{2}{3}x=\frac{55}{81}$$

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{55}{81}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{55}{81}+\frac{1}{9}$$

Add \frac{55}{81} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{64}{81}$$

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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{64}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{64}{81}}$$

Simplify.

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

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

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