Question

what values of b satisfy 4(3b + 2)² = 64?
○ b = 2/3 and b = -2
○ b = 2 and b = 10/3
○ b = 2/3 and b = 3
○ b = 2 and b = -10/3

Explanation:

Step1: Divide both sides by 4

To isolate the squared term, divide each side of the equation \(4(3b + 2)^2 = 64\) by 4.
\(\frac{4(3b + 2)^2}{4} = \frac{64}{4}\)
Simplifying gives \((3b + 2)^2 = 16\).

Step2: Take square root of both sides

Take the square root of both sides. Remember that taking the square root of a number gives two solutions: positive and negative.
\(\sqrt{(3b + 2)^2} = \pm\sqrt{16}\)
This simplifies to \(3b + 2 = \pm 4\).

Step3: Solve for \(b\) in two cases

Case 1: \(3b + 2 = 4\)

Subtract 2 from both sides: \(3b = 4 - 2\)
\(3b = 2\)
Divide by 3: \(b = \frac{2}{3}\)

Case 2: \(3b + 2 = -4\)

Subtract 2 from both sides: \(3b = -4 - 2\)
\(3b = -6\)
Divide by 3: \(b = -2\)

Answer:

\(b = \frac{2}{3}\) and \(b = -2\) (the first option: \(b = \frac{2}{3}\) and \(b = -2\))