What values of b satisfy 4(3b + 2)2 = 64?

a. b = 2/3 and b = –2

b. b = 2 and b = 10/3

c. b = 2/3 and b = 3

d. b = 2 and b = -10/3

Given the equation 4(3b + 2)² = 64, dividing both sides of the equation by 4, we have (3b + 2)² = 16 and getting the square root of both sides, (3b + 2) = 4 and (3b + 2) = -4 We can solve for b for each equation and have 3b = 2 | 3b = -6 b = 2/3 | b = -2 Therefore, the values of b are 2/3 and -2 and from the choices, the answer is A: b = 2/3 and b = -2.