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The equation given is \( b^2 = 49 \). To solve this, we need to find the values of \( b \) that satisfy this equation.

### Solution:

1. Take the square root of both sides:
   \[
   b = \pm \sqrt{49}
   \]
   Since the square of both positive and negative values can equal 49, we have two possible solutions:
   \[
   b = 7 \quad \text{or} \quad b = -7
   \]

Now let's evaluate the options:
- **A)** \( b = 7 \) is correct.
- **B)** \( b = -7 \) is also correct.
- **C)** \( b = \sqrt{49} \) simplifies to \( b = 7 \), so this is also correct.
- **D)** \( b = -\sqrt{49} \) simplifies to \( b = -7 \), so this is correct as well.
- **E)** "None of the above" is incorrect since multiple answers are valid.

Thus, the correct answers are **A, B, C, D**.

Would you like more details or have any questions? 

Here are five related questions to explore:
1. What is the significance of both positive and negative solutions when solving square equations?
2. How do we handle square root equations with complex numbers?
3. What happens when the number under the square root is negative?
4. Can square root operations be applied to non-integer values?
5. How can this concept be extended to higher powers, like \( b^3 = 49 \)?

**Tip:** Always remember to include both positive and negative roots when solving equations like \( x^2 = n \).

Find the solution(s) of the equation b^2 = 49 and choose all the correct answers.

Learn how to solve the quadratic equation b^2 = 49 and understand why the solutions include both positive and negative square roots. This problem covers fundamental algebraic concepts like square roots and quadratic equations, suitable for students in Grades 7-10.

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