Question

what values of b satisfy $3(2b + 3)^2 = 36$? $b = \frac{-3 + 2\sqrt{3}}{3}$ and $\frac{-3 - 2\sqrt{3}}{3}$; $b = \frac{-3 + 2\sqrt{3}}{2}$ and $\frac{-3 - 2\sqrt{3}}{2}$; $b = \frac{9}{2}$ and $-\frac{3}{2}$

Explanation:

Step1: Divide both sides by 3

$\frac{3(2b+3)^2}{3} = \frac{36}{3}$
$(2b+3)^2 = 12$

Step2: Take square root of both sides

$2b+3 = \pm\sqrt{12} = \pm2\sqrt{3}$

Step3: Isolate the term with b

$2b = -3 \pm2\sqrt{3}$

Step4: Solve for b

$b = \frac{-3 \pm2\sqrt{3}}{2}$

Answer:

$b = \frac{-3+2\sqrt{3}}{2}$ and $\frac{-3-2\sqrt{3}}{2}$