https://file.aiquickdraw.com/mathbot/file/1737454217677801309t7.jpg

Let's solve each inequality step by step:

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### **(f) \( -x^2 - 5x - 6 \geq 0 \)**

1. Rewrite it in standard quadratic form:
   \[
   x^2 + 5x + 6 \leq 0
   \]

2. Factorize the quadratic expression:
   \[
   x^2 + 5x + 6 = (x + 2)(x + 3)
   \]

3. Solve the equality \( (x + 2)(x + 3) = 0 \), which gives:
   \[
   x = -2 \quad \text{and} \quad x = -3
   \]

4. Use these roots to divide the real number line into intervals:
   \[
   (-\infty, -3), \ (-3, -2), \ (-2, \infty)
   \]

5. Test the sign of \( (x + 2)(x + 3) \) in each interval:
   - For \( x \in (-\infty, -3) \), \( (x + 2)(x + 3) > 0 \).
   - For \( x \in (-3, -2) \), \( (x + 2)(x + 3) < 0 \).
   - For \( x \in (-2, \infty) \), \( (x + 2)(x + 3) > 0 \).

6. Since \( x^2 + 5x + 6 \leq 0 \), the solution is:
   \[
   x \in [-3, -2]
   \]

---

### **(i) \( x(x - 1) \leq 20 \)**

1. Rewrite as:
   \[
   x^2 - x - 20 \leq 0
   \]

2. Factorize the quadratic expression:
   \[
   x^2 - x - 20 = (x - 5)(x + 4)
   \]

3. Solve the equality \( (x - 5)(x + 4) = 0 \), which gives:
   \[
   x = 5 \quad \text{and} \quad x = -4
   \]

4. Use these roots to divide the real number line into intervals:
   \[
   (-\infty, -4), \ (-4, 5), \ (5, \infty)
   \]

5. Test the sign of \( (x - 5)(x + 4) \) in each interval:
   - For \( x \in (-\infty, -4) \), \( (x - 5)(x + 4) > 0 \).
   - For \( x \in (-4, 5) \), \( (x - 5)(x + 4) < 0 \).
   - For \( x \in (5, \infty) \), \( (x - 5)(x + 4) > 0 \).

6. Since \( x^2 - x - 20 \leq 0 \), the solution is:
   \[
   x \in [-4, 5]
   \]

---

### **(l) \( x^2 + 3x \geq -2 \)**

1. Rewrite as:
   \[
   x^2 + 3x + 2 \geq 0
   \]

2. Factorize the quadratic expression:
   \[
   x^2 + 3x + 2 = (x + 1)(x + 2)
   \]

3. Solve the equality \( (x + 1)(x + 2) = 0 \), which gives:
   \[
   x = -1 \quad \text{and} \quad x = -2
   \]

4. Use these roots to divide the real number line into intervals:
   \[
   (-\infty, -2), \ (-2, -1), \ (-1, \infty)
   \]

5. Test the sign of \( (x + 1)(x + 2) \) in each interval:
   - For \( x \in (-\infty, -2) \), \( (x + 1)(x + 2) > 0 \).
   - For \( x \in (-2, -1) \), \( (x + 1)(x + 2) < 0 \).
   - For \( x \in (-1, \infty) \), \( (x + 1)(x + 2) > 0 \).

6. Since \( x^2 + 3x + 2 \geq 0 \), the solution is:
   \[
   x \in (-\infty, -2] \cup [-1, \infty)
   \]

---

### Final Solutions:
- **(f)**: \( x \in [-3, -2] \)
- **(i)**: \( x \in [-4, 5] \)
- **(l)**: \( x \in (-\infty, -2] \cup [-1, \infty) \)

Let me know if you'd like further details!

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### Related Questions:
1. What is the method to graph solutions for quadratic inequalities?
2. How do test points help in determining the intervals of a quadratic inequality?
3. What does the sign of a quadratic expression depend on?
4. Why is the factorization method essential in solving quadratic inequalities?
5. How would the solution change for strict inequalities (e.g., \( < \) or \( > \))?

### Tip:
Always double-check the direction of the inequality after testing intervals to ensure correct results.

Solve the inequalities provided in the image: (f) -x^2 - 5x - 6 ≥ 0, (i) x(x - 1) ≤ 20, and (l) x^2 + 3x ≥ -2.

Learn how to solve quadratic inequalities including -x^2 - 5x - 6 ≥ 0, x(x - 1) ≤ 20, and x^2 + 3x ≥ -2 with detailed explanations and step-by-step methods. Covers concepts like factorization, testing intervals, and quadratic inequality solutions. Perfect for Grades 9-11 students.

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