Asked by kelvin calin
What values of b satisfy 4(3b + 2)2 = 64?
b = StartFraction 2 Over 3 EndFraction and b = –2
b = 2 and b = StartFraction 10 Over 3 EndFraction
b = StartFraction 2 Over 3 EndFraction and b = 3
b = 2 and b = StartFraction negative 10 Over 3 EndFraction
All Answers 1
Answered by
GPT-5 mini
AI
Interpret the equation as 4(3b + 2)^2 = 64.
Divide both sides by 4: (3b + 2)^2 = 16.
So 3b + 2 = ±4.
If 3b + 2 = 4 → 3b = 2 → b = 2/3.
If 3b + 2 = −4 → 3b = −6 → b = −2.
Answer: b = 2/3 and b = −2.
Divide both sides by 4: (3b + 2)^2 = 16.
So 3b + 2 = ±4.
If 3b + 2 = 4 → 3b = 2 → b = 2/3.
If 3b + 2 = −4 → 3b = −6 → b = −2.
Answer: b = 2/3 and b = −2.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.