Question

TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?

Answer 1

Answer:

$\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}$

Step-by-step explanation:

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

Taking square root both sides, we get

$\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}$

Therefore, the values of $b$ are:

$\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}$

Answer 2

Answer:

A

Step-by-step explanation: