Question

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

Answer 1

Answer:

A. b = StartFraction 2 Over 3 EndFraction and b = –2

Answer 2

Assuming yo meant
4(3b+2)²=64

divide both sides by 4
(3b+2)²=16
sqrt both sides
remember to take positive and negative roots
3b+2=+/-4
minus 2 both sides
3b=-2+/-4
divide by 3
b=(-2+/-4)/3
b=(-2+4)/3 or (-2-4)/3
b=2/3 or -6/3
b=2/3 or -2


A is answer