Question

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

Answer 1

b=\frac{2}{3} and b=-2

Explanation:

Answer 2

`b=(2)/(3)` and
`b=-2`

Explanation:

Given:

`4(3b+2)^(2)=64`

In order to find the values of b, we need to solve the equation for b.

Divide both sides by 4.

This gives,

`4(3b+2)^(2)=64\\ (4(3b+2)^(2))/(4)=(64)/(4)\\ (3b+2)^(2)=16`

Taking square root both sides, we get,

`√((3b+2)^2)=√(16)\\3b+2=\pm 4\\3b=\pm 4-2\\b=(-2\pm 4)/(3)`

Now,
`b=(-2+4)/(3)\textrm{ or } b=(-2-4)/(3)\\b=(2)/(3)\textrm{ or } b=(-6)/(3)\\b=(2)/(3)\textrm{ or } b=-2`