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

*b* =  and *b* = –2

*b* = 2 and *b* =

*b* =  and *b* = 3

*b* = 2 and *b* =

We are given the equation:

\[
4(3b + 2)^2 = 64
\]

To solve for \( b \), let's break it down step by step:

### Step 1: Simplify the equation
Divide both sides of the equation by 4:

\[
(3b + 2)^2 = \frac{64}{4} = 16
\]

### Step 2: Take the square root of both sides
Now, take the square root of both sides of the equation:

\[
3b + 2 = \pm \sqrt{16}
\]

\[
3b + 2 = \pm 4
\]

### Step 3: Solve for \( b \)
We now have two cases to solve:

1. \( 3b + 2 = 4 \)
2. \( 3b + 2 = -4 \)

#### Case 1: \( 3b + 2 = 4 \)
Subtract 2 from both sides:

\[
3b = 4 - 2 = 2
\]

Now, divide by 3:

\[
b = \frac{2}{3}
\]

#### Case 2: \( 3b + 2 = -4 \)
Subtract 2 from both sides:

\[
3b = -4 - 2 = -6
\]

Now, divide by 3:

\[
b = \frac{-6}{3} = -2
\]

### Final Answer:
The values of \( b \) are \( b = \frac{2}{3} \) and \( b = -2 \).

Let me know if you would like any further details or clarifications!

Here are some related questions:
1. How do you solve quadratic equations in general?
2. What happens when you take the square root of a negative number?
3. How do you check if your solutions for \( b \) are correct?
4. Can there be more than two solutions to this type of equation?
5. How does squaring affect the outcome of equations like these?

**Tip:** When solving equations involving squares, always remember to consider both the positive and negative square roots!

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

Discover how to solve the quadratic equation 4(3b + 2)^2 = 64 step-by-step, using algebraic concepts such as expanding squares and applying square root principles. Ideal for students in Grades 8-10.

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