To solve the inequality:

\[
(2x + 1)^2 > 36
\]

1. **Take the square root of both sides:**  
   Since \((2x + 1)^2 > 36\), we can take the square root on both sides, remembering that this leads to two cases (positive and negative roots):

   \[
   |2x + 1| > 6
   \]

2. **Set up two inequalities:**  
   This absolute value inequality splits into two inequalities:

   \[
   2x + 1 > 6 \quad \text{or} \quad 2x + 1 < -6
   \]

3. **Solve each inequality:**

   - For \(2x + 1 > 6\):
     \[
     2x > 5
     \]
     \[
     x > \frac{5}{2} \quad \text{or} \quad x > 2.5
     \]

   - For \(2x + 1 < -6\):
     \[
     2x < -7
     \]
     \[
     x < -\frac{7}{2} \quad \text{or} \quad x < -3.5
     \]

4. **Combine the solutions:**
   The solution is:

   \[
   x > 2.5 \quad \text{or} \quad x < -3.5
   \]

In interval notation, this solution is:

\[
(-\infty, -3.5) \cup (2.5, \infty)
\]

Would you like more details on any part of this process?

### Related Questions
1. What if the inequality was \((2x + 1)^2 \leq 36\)?
2. How would we solve if it was \((3x + 2)^2 > 49\)?
3. What happens if we change \(2x + 1\) to another linear expression?
4. How can we solve inequalities involving cubes, such as \((3x - 4)^3 > 27\)?
5. What if we needed to solve \((2x + 1)^2 - 36 > 0\) instead?

### Tip
When solving inequalities with squares, always consider both positive and negative square root cases.

Explore a step-by-step solution to the quadratic inequality (2x + 1)² > 36. Learn how to apply absolute value principles and solve inequalities by splitting cases. Suitable for students in Grades 9-10, covering key concepts in algebra and inequality-solving techniques.

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