Solving Quadratic Inequalities worksheet with eight problems to solve.

Let's solve each of the quadratic inequalities step by step. We'll follow a standard method:

General Strategy for Solving Quadratic Inequalities:

1. Solve the corresponding quadratic equation (set inequality to zero) to find critical points.
2. Plot the critical points on a number line to divide it into intervals.
3. Test a point in each interval to determine where the inequality holds.
4. Include endpoints if the inequality is non-strict (≤ or ≥).

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1) $ x^2 - 9x + 20 > 0 $

Step 1: Solve $ x^2 - 9x + 20 = 0 $

Factor:
$$
x^2 - 9x + 20 = (x - 4)(x - 5) = 0
\Rightarrow x = 4, \; x = 5
$$

Step 2: Critical points: $ x = 4, 5 $. These divide the number line into three intervals:
- $ (-\infty, 4) $
- $ (4, 5) $
- $ (5, \infty) $

Step 3: Test each interval:

- Pick $ x = 0 $: $ (0 - 4)(0 - 5) = (-4)(-5) = 20 > 0 $ → satisfies
- Pick $ x = 4.5 $: $ (4.5 - 4)(4.5 - 5) = (0.5)(-0.5) = -0.25 < 0 $ → does not satisfy
- Pick $ x = 6 $: $ (6 - 4)(6 - 5) = (2)(1) = 2 > 0 $ → satisfies

Step 4: Since inequality is > 0, we want positive values.

So solution:
$$
x < 4 \quad \text{or} \quad x > 5
$$
$$
\boxed{(-\infty, 4) \cup (5, \infty)}
$$

---

2) $ m^2 + 7m - 18 < 0 $

Step 1: Solve $ m^2 + 7m - 18 = 0 $

Factor:
$$
m^2 + 7m - 18 = (m + 9)(m - 2) = 0
\Rightarrow m = -9, \; m = 2
$$

Step 2: Intervals: $ (-\infty, -9), (-9, 2), (2, \infty) $

Step 3: Test:

- $ m = -10 $: $ (-10+9)(-10-2) = (-1)(-12) = 12 > 0 $
- $ m = 0 $: $ (0+9)(0-2) = (9)(-2) = -18 < 0 $ → satisfies
- $ m = 3 $: $ (3+9)(3-2) = (12)(1) = 12 > 0 $

We want < 0, so only middle interval works.

Solution:
$$
\boxed{(-9, 2)}
$$

---

3) $ z^2 - 9 \leq 0 $

This is a difference of squares:
$$
z^2 - 9 = (z - 3)(z + 3) \leq 0
$$

Critical points: $ z = -3, 3 $

Intervals: $ (-\infty, -3), (-3, 3), (3, \infty) $

Test:

- $ z = -4 $: $ (-4-3)(-4+3) = (-7)(-1) = 7 > 0 $
- $ z = 0 $: $ (-3)(3) = -9 < 0 $
- $ z = 4 $: $ (1)(7) = 7 > 0 $

We want ≤ 0 → includes equality.

So solution: $ [-3, 3] $

$$
\boxed{[-3, 3]}
$$

---

4) $ p^2 - 4p - 12 \geq 0 $

Solve $ p^2 - 4p - 12 = 0 $

Factor:
$$
(p - 6)(p + 2) = 0 \Rightarrow p = -2, 6
$$

Intervals: $ (-\infty, -2), (-2, 6), (6, \infty) $

Test:

- $ p = -3 $: $ (-3-6)(-3+2) = (-9)(-1) = 9 > 0 $
- $ p = 0 $: $ (-6)(2) = -12 < 0 $
- $ p = 7 $: $ (1)(9) = 9 > 0 $

We want $ \geq 0 $ → includes endpoints.

So solution: $ (-\infty, -2] \cup [6, \infty) $

$$
\boxed{(-\infty, -2] \cup [6, \infty)}
$$

---

5) $ r^2 + 3r - 18 < 0 $

Solve $ r^2 + 3r - 18 = 0 $

Factor:
$$
(r + 6)(r - 3) = 0 \Rightarrow r = -6, 3
$$

Intervals: $ (-\infty, -6), (-6, 3), (3, \infty) $

Test:

- $ r = -7 $: $ (-7+6)(-7-3) = (-1)(-10) = 10 > 0 $
- $ r = 0 $: $ (6)(-3) = -18 < 0 $
- $ r = 4 $: $ (10)(1) = 10 > 0 $

We want < 0 → only middle interval.

$$
\boxed{(-6, 3)}
$$

---

6) $ n^2 - 16 > 0 $

Difference of squares:
$$
n^2 - 16 = (n - 4)(n + 4) > 0
$$

Critical points: $ n = -4, 4 $

Intervals: $ (-\infty, -4), (-4, 4), (4, \infty) $

Test:

- $ n = -5 $: $ (-9)(-1) = 9 > 0 $
- $ n = 0 $: $ (-4)(4) = -16 < 0 $
- $ n = 5 $: $ (1)(9) = 9 > 0 $

Want > 0 → outer intervals.

$$
\boxed{(-\infty, -4) \cup (4, \infty)}
$$

---

7) $ q^2 - 10q - 11 \geq 0 $

Solve $ q^2 - 10q - 11 = 0 $

Use quadratic formula:
$$
q = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-11)}}{2} = \frac{10 \pm \sqrt{100 + 44}}{2} = \frac{10 \pm \sqrt{144}}{2} = \frac{10 \pm 12}{2}
$$
$$
q = \frac{22}{2} = 11, \quad q = \frac{-2}{2} = -1
$$

So roots: $ q = -1, 11 $

Intervals: $ (-\infty, -1), (-1, 11), (11, \infty) $

Test:

- $ q = -2 $: $ (-2)^2 -10(-2) -11 = 4 + 20 -11 = 13 > 0 $
- $ q = 0 $: $ 0 - 0 -11 = -11 < 0 $
- $ q = 12 $: $ 144 - 120 -11 = 13 > 0 $

We want $ \geq 0 $ → include endpoints.

So solution: $ (-\infty, -1] \cup [11, \infty) $

$$
\boxed{(-\infty, -1] \cup [11, \infty)}
$$

---

8) $ a^2 + 8a + 15 \leq 0 $

Factor:
$$
a^2 + 8a + 15 = (a + 3)(a + 5) \leq 0
$$

Roots: $ a = -5, -3 $

Intervals: $ (-\infty, -5), (-5, -3), (-3, \infty) $

Test:

- $ a = -6 $: $ (-6+3)(-6+5) = (-3)(-1) = 3 > 0 $
- $ a = -4 $: $ (-1)(1) = -1 < 0 $
- $ a = 0 $: $ (3)(5) = 15 > 0 $

We want ≤ 0 → middle interval including endpoints.

$$
\boxed{[-5, -3]}
$$

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✔ Final Answers:

1) $ \boxed{(-\infty, 4) \cup (5, \infty)} $
2) $ \boxed{(-9, 2)} $
3) $ \boxed{[-3, 3]} $
4) $ \boxed{(-\infty, -2] \cup [6, \infty)} $
5) $ \boxed{(-6, 3)} $
6) $ \boxed{(-\infty, -4) \cup (4, \infty)} $
7) $ \boxed{(-\infty, -1] \cup [11, \infty)} $
8) $ \boxed{[-5, -3]} $

Let me know if you'd like a graph or explanation for any one!