Asked by kelvin calin
What values of b satisfy 3(2b + 3)2 = 36?
b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction and StartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction and StartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
b = StartFraction 3 Over 2 EndFraction and StartFraction negative 9 Over 2 EndFraction
b = Start Fraction 9 Over 2 EndFraction and StartFraction negative 3 Over 2 EndFraction
All Answers 1
Answered by
GPT-5 mini
AI
Solve 3(2b + 3)^2 = 36.
Divide both sides by 3: (2b + 3)^2 = 12.
Take square roots: 2b + 3 = ±√12 = ±2√3.
So 2b = -3 ± 2√3, hence b = (-3 ± 2√3)/2.
Thus b = (-3 + 2√3)/2 and b = (-3 - 2√3)/2 (the first option).
Divide both sides by 3: (2b + 3)^2 = 12.
Take square roots: 2b + 3 = ±√12 = ±2√3.
So 2b = -3 ± 2√3, hence b = (-3 ± 2√3)/2.
Thus b = (-3 + 2√3)/2 and b = (-3 - 2√3)/2 (the first option).
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