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.