Solve (3k+4)^2=25

$$( 3 k + 4 ) ^ { 2 } = 25$$

Solve for k

$$k=\frac{1}{3}\approx 0.333333333$$

$$k=-3$$

Steps Using Factoring By Grouping

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

$$9k^{2}+24k+16=25$$

Subtract 25 from both sides.

$$9k^{2}+24k+16-25=0$$

Subtract 25 from 16 to get -9.

$$9k^{2}+24k-9=0$$

Divide both sides by 3.

$$3k^{2}+8k-3=0$$

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

$$a+b=8 ab=3\left(-3\right)=-9$$

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 -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=-1 b=9$$

Rewrite 3k^{2}+8k-3 as \left(3k^{2}-k\right)+\left(9k-3\right).

$$\left(3k^{2}-k\right)+\left(9k-3\right)$$

Factor out k in the first and 3 in the second group.

$$k\left(3k-1\right)+3\left(3k-1\right)$$

Factor out common term 3k-1 by using distributive property.

$$\left(3k-1\right)\left(k+3\right)$$

To find equation solutions, solve 3k-1=0 and k+3=0.

$$k=\frac{1}{3} k=-3$$

Steps Using the Quadratic Formula

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

$$9k^{2}+24k+16=25$$

Subtract 25 from both sides.

$$9k^{2}+24k+16-25=0$$

Subtract 25 from 16 to get -9.

$$9k^{2}+24k-9=0$$

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

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

Square 24.

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

Multiply -4 times 9.

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

Multiply -36 times -9.

$$k=\frac{-24±\sqrt{576+324}}{2\times 9}$$

Add 576 to 324.

$$k=\frac{-24±\sqrt{900}}{2\times 9}$$

Take the square root of 900.

$$k=\frac{-24±30}{2\times 9}$$

Multiply 2 times 9.

$$k=\frac{-24±30}{18}$$

Now solve the equation k=\frac{-24±30}{18} when ± is plus. Add -24 to 30.

$$k=\frac{6}{18}$$

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

$$k=\frac{1}{3}$$

Now solve the equation k=\frac{-24±30}{18} when ± is minus. Subtract 30 from -24.

$$k=-\frac{54}{18}$$

Divide -54 by 18.

$$k=-3$$

The equation is now solved.

$$k=\frac{1}{3} k=-3$$

Steps for Completing the Square

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

$$9k^{2}+24k+16=25$$

Subtract 16 from both sides.

$$9k^{2}+24k=25-16$$

Subtract 16 from 25 to get 9.

$$9k^{2}+24k=9$$

Divide both sides by 9.

$$\frac{9k^{2}+24k}{9}=\frac{9}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$k^{2}+\frac{24}{9}k=\frac{9}{9}$$

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

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

Divide 9 by 9.

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

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.

$$k^{2}+\frac{8}{3}k+\left(\frac{4}{3}\right)^{2}=1+\left(\frac{4}{3}\right)^{2}$$

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

$$k^{2}+\frac{8}{3}k+\frac{16}{9}=1+\frac{16}{9}$$

Add 1 to \frac{16}{9}.

$$k^{2}+\frac{8}{3}k+\frac{16}{9}=\frac{25}{9}$$

Factor k^{2}+\frac{8}{3}k+\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}.

$$\left(k+\frac{4}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{4}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$k+\frac{4}{3}=\frac{5}{3} k+\frac{4}{3}=-\frac{5}{3}$$

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

$$k=\frac{1}{3} k=-3$$