Question

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

Explanation:

Step1: Divide both sides by 3

To isolate the squared term, we divide both sides of the equation \(3(2b + 3)^2 = 36\) by 3.
\(\frac{3(2b + 3)^2}{3}=\frac{36}{3}\)
Simplifying, we get \((2b + 3)^2 = 12\).

Step2: Take square root of both sides

Taking the square root of both sides, we have \(2b + 3=\pm\sqrt{12}\).
Simplify \(\sqrt{12}\) as \(2\sqrt{3}\), so \(2b + 3=\pm2\sqrt{3}\).

Step3: Solve for b

First, subtract 3 from both sides: \(2b=-3\pm2\sqrt{3}\).
Then, divide both sides by 2: \(b = \frac{-3\pm2\sqrt{3}}{2}\).

Answer:

\(b=\frac{-3 + 2\sqrt{3}}{2}\) and \(\frac{-3 - 2\sqrt{3}}{2}\) (the first option)