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Let's solve the quadratic inequalities one by one, as presented on the board:

### 1. \( x^2 + 5x - 24 > 0 \)

We need to solve this quadratic inequality. First, let's find the roots by solving the equation:
\[
x^2 + 5x - 24 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 5 \), and \( c = -24 \).

\[
x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-24)}}{2(1)}
\]
\[
x = \frac{-5 \pm \sqrt{25 + 96}}{2}
\]
\[
x = \frac{-5 \pm \sqrt{121}}{2}
\]
\[
x = \frac{-5 \pm 11}{2}
\]

Thus, the two solutions are:
\[
x_1 = \frac{-5 + 11}{2} = 3, \quad x_2 = \frac{-5 - 11}{2} = -8
\]

Now, we can break the number line into three intervals based on the roots:
- \( (-\infty, -8) \)
- \( (-8, 3) \)
- \( (3, \infty) \)

We test points from each interval in the original inequality \( x^2 + 5x - 24 > 0 \):
- For \( x = -9 \) (from \( (-\infty, -8) \)): \( (-9)^2 + 5(-9) - 24 = 81 - 45 - 24 = 12 \), which is positive.
- For \( x = 0 \) (from \( (-8, 3) \)): \( (0)^2 + 5(0) - 24 = -24 \), which is negative.
- For \( x = 4 \) (from \( (3, \infty) \)): \( (4)^2 + 5(4) - 24 = 16 + 20 - 24 = 12 \), which is positive.

Thus, the solution for \( x^2 + 5x - 24 > 0 \) is:
\[
x \in (-\infty, -8) \cup (3, \infty)
\]

### 2. \( x^2 + 6x \leq 0 \)

We start by factoring the quadratic expression:
\[
x^2 + 6x = x(x + 6)
\]
Now solve the corresponding equation:
\[
x(x + 6) = 0
\]
So the roots are \( x = 0 \) and \( x = -6 \).

We break the number line into intervals:
- \( (-\infty, -6) \)
- \( (-6, 0) \)
- \( (0, \infty) \)

Now test points from each interval:
- For \( x = -7 \) (from \( (-\infty, -6) \)): \( (-7)(-7 + 6) = (-7)(-1) = 7 \), which is positive.
- For \( x = -3 \) (from \( (-6, 0) \)): \( (-3)(-3 + 6) = (-3)(3) = -9 \), which is negative.
- For \( x = 1 \) (from \( (0, \infty) \)): \( (1)(1 + 6) = (1)(7) = 7 \), which is positive.

Thus, the inequality holds when \( x \in [-6, 0] \).

### 3. \( 2x^2 - 72 \geq 0 \)

We start by simplifying the inequality:
\[
2(x^2 - 36) \geq 0
\]
\[
x^2 - 36 \geq 0
\]
\[
(x - 6)(x + 6) \geq 0
\]

The roots are \( x = 6 \) and \( x = -6 \).

We break the number line into intervals:
- \( (-\infty, -6) \)
- \( (-6, 6) \)
- \( (6, \infty) \)

Now test points from each interval:
- For \( x = -7 \) (from \( (-\infty, -6) \)): \( (-7)^2 - 36 = 49 - 36 = 13 \), which is positive.
- For \( x = 0 \) (from \( (-6, 6) \)): \( (0)^2 - 36 = -36 \), which is negative.
- For \( x = 7 \) (from \( (6, \infty) \)): \( (7)^2 - 36 = 49 - 36 = 13 \), which is positive.

Thus, the solution for \( 2x^2 - 72 \geq 0 \) is:
\[
x \in (-\infty, -6] \cup [6, \infty)
\]

---

Would you like further details or clarifications on any part? Here are some follow-up questions:

1. How do you determine intervals when solving quadratic inequalities?
2. Why is testing points in each interval important?
3. Can quadratic inequalities have solutions in closed intervals?
4. How do you factor quadratics to solve inequalities?
5. How would the approach change if the inequalities were strict (e.g., \( < \) or \( > \))?

**Tip:** Always check your factored expressions carefully and ensure you test points properly from all intervals when solving inequalities.

Find the solution set of the following quadratic inequality:
1. x^2 + 5x - 24 > 0
2. x^2 + 6x ≤ 0
3. 2x^2 - 72 ≥ 0

Learn how to solve the quadratic inequalities x^2 + 5x - 24 > 0, x^2 + 6x ≤ 0, and 2x^2 - 72 ≥ 0 with detailed step-by-step solutions. This problem covers key algebraic concepts, including solving quadratic equations and analyzing the sign of quadratic expressions. Suitable for students in Grades 10-12.

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