Solve left(3b+4right)^2-64=0

$${ \left(3b+4 \right) }^{ 2 } -64 = 0$$

Solve for b

$$b = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$b=-4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3b+4\right)^{2}.

$$9b^{2}+24b+16-64=0$$

Subtract 64 from 16 to get -48.

$$9b^{2}+24b-48=0$$

Divide both sides by 3.

$$3b^{2}+8b-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3b^{2}+ab+bb-16. To find a and b, set up a system to be solved.

$$a+b=8 ab=3\left(-16\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=-4 b=12$$

Rewrite 3b^{2}+8b-16 as \left(3b^{2}-4b\right)+\left(12b-16\right).

$$\left(3b^{2}-4b\right)+\left(12b-16\right)$$

Factor out b in the first and 4 in the second group.

$$b\left(3b-4\right)+4\left(3b-4\right)$$

Factor out common term 3b-4 by using distributive property.

$$\left(3b-4\right)\left(b+4\right)$$

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3b+4\right)^{2}.

$$9b^{2}+24b+16-64=0$$

Subtract 64 from 16 to get -48.

$$9b^{2}+24b-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 24 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-24±\sqrt{24^{2}-4\times 9\left(-48\right)}}{2\times 9}$$

Square 24.

$$b=\frac{-24±\sqrt{576-4\times 9\left(-48\right)}}{2\times 9}$$

Multiply -4 times 9.

$$b=\frac{-24±\sqrt{576-36\left(-48\right)}}{2\times 9}$$

Multiply -36 times -48.

$$b=\frac{-24±\sqrt{576+1728}}{2\times 9}$$

Add 576 to 1728.

$$b=\frac{-24±\sqrt{2304}}{2\times 9}$$

Take the square root of 2304.

$$b=\frac{-24±48}{2\times 9}$$

Multiply 2 times 9.

$$b=\frac{-24±48}{18}$$

Now solve the equation b=\frac{-24±48}{18} when ± is plus. Add -24 to 48.

$$b=\frac{24}{18}$$

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

$$b=\frac{4}{3}$$

Now solve the equation b=\frac{-24±48}{18} when ± is minus. Subtract 48 from -24.

$$b=-\frac{72}{18}$$

Divide -72 by 18.

$$b=-4$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3b+4\right)^{2}.

$$9b^{2}+24b+16-64=0$$

Subtract 64 from 16 to get -48.

$$9b^{2}+24b-48=0$$

Add 48 to both sides. Anything plus zero gives itself.

$$9b^{2}+24b=48$$

Divide both sides by 9.

$$\frac{9b^{2}+24b}{9}=\frac{48}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$b^{2}+\frac{24}{9}b=\frac{48}{9}$$

Reduce the fraction \frac{24}{9} to lowest terms by extracting and canceling out 3.

$$b^{2}+\frac{8}{3}b=\frac{48}{9}$$

Reduce the fraction \frac{48}{9} to lowest terms by extracting and canceling out 3.

$$b^{2}+\frac{8}{3}b=\frac{16}{3}$$

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+\frac{8}{3}b+\left(\frac{4}{3}\right)^{2}=\frac{16}{3}+\left(\frac{4}{3}\right)^{2}$$

Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$b^{2}+\frac{8}{3}b+\frac{16}{9}=\frac{16}{3}+\frac{16}{9}$$

Add \frac{16}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}+\frac{8}{3}b+\frac{16}{9}=\frac{64}{9}$$

Factor b^{2}+\frac{8}{3}b+\frac{16}{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}.

$$\left(b+\frac{4}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{4}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$b+\frac{4}{3}=\frac{8}{3} b+\frac{4}{3}=-\frac{8}{3}$$

Subtract \frac{4}{3} from both sides of the equation.

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