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

Respuesta :

Answer: b=-3+2√3/2

-3-2√3/2

Step-by-step explanation: it's A on edg

Lanuel

The value of b which satisfies the given quadratic equation is both: A. b = (-3 + 2√3)/2 and (-3 - 2√3)/2.

How to solve a quadratic equation?

In this exercise, you're required to solve for the value of b which satisfies the given quadratic equation. Thus, we would simplify the quadratic equation as follows:

3(2b + 3)² = 36

Dividing both sides by 3, we have:

(2b + 3)² = 12

Taking the square root of both sides, we have:

2b + 3 = ±√12

2b + 3 = ±√4 × √3

2b + 3 = ±2√3

2b = -3 ± 2√3

b = (-3 + 2√3)/2    and    (-3 - 2√3)/2.  

Read more on quadratic equation here: brainly.com/question/1214333

#SPJ2