Question

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

a. b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction andStartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
b. b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction and StartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
c. b = StartFraction 3 Over 2 EndFraction and StartFraction negative 9 Over 2 EndFraction
d. b = Start Fraction 9 Over 2 EndFraction and StartFraction negative 3 Over 2 EndFraction

Answer 1

A

Explanation:

Answer 2

`(-3+ 2√(3))/(2)\textrm{ and }(-3- 2√(3))/(2)`

Explanation:

`3(2b+3)^(2)=36\\(3(2b+3)^(2))/(3)=(36)/(3)\\(2b+3)^(2)=12`

Taking square root both sides, we get

`\sqrt{(2b+3)^(2)}=\pm √(12)\\2b+3=\pm √(4* 3)\\2b+3=\pm 2√(3)\\2b+3-3=\pm 2√(3)-3\\2b=-3\pm 2√(3)\\b=(-3\pm 2√(3))/(2)\\b=(-3+ 2√(3))/(2)\textrm{ or }b=(-3- 2√(3))/(2)`

Therefore, the values of
`b` are:

`(-3+ 2√(3))/(2)\textrm{ and }(-3- 2√(3))/(2)`