Quadratic Inequalities Worksheet for 9th Grade | Lesson Planet - Free Printable

Problem Analysis:

The task involves solving quadratic inequalities and determining the solution sets. The goal is to find the values of \( x \) that satisfy each inequality and then represent the solution set on a number line.

#### Quadratic Inequalities:
1. \( x^2 - 8x = 0 \)
2. \( x^2 - x + 20 \geq 0 \)
3. \( x^2 - 4x - 24 < 0 \)
4. \( x^2 + 8x \geq 0 \)
5. \( x^2 - 6x - 16 \geq 0 \)
6. \( x^2 - 14x > 0 \)
7. \( x^2 - 21x + 108 \leq 0 \)
8. \( x^2 + 4x + 5 \geq 0 \)

Additionally, there are two compound inequalities:
9. \( y \leq x^2 - 8x + 16 \)
10. \( y \geq x^2 + 4x - 6 \)

---

Step-by-Step Solution:

#### 1. Solve \( x^2 - 8x = 0 \)

Factorize:
\[
x(x - 8) = 0
\]
Solutions:
\[
x = 0 \quad \text{or} \quad x = 8
\]

This is not an inequality, so the solution is:
\[
\boxed{x = 0 \text{ or } x = 8}
\]

---

#### 2. Solve \( x^2 - x + 20 \geq 0 \)

First, solve the corresponding equation:
\[
x^2 - x + 20 = 0
\]
Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, b = -1, c = 20
\]
\[
x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(20)}}{2(1)} = \frac{1 \pm \sqrt{1 - 80}}{2} = \frac{1 \pm \sqrt{-79}}{2}
\]
Since the discriminant (\( -79 \)) is negative, there are no real roots. The parabola \( x^2 - x + 20 \) opens upwards (since the coefficient of \( x^2 \) is positive), and it never touches the x-axis. Therefore, \( x^2 - x + 20 \) is always positive for all real \( x \).

Solution:
\[
\boxed{(-\infty, \infty)}
\]

---

#### 3. Solve \( x^2 - 4x - 24 < 0 \)

Factorize:
\[
x^2 - 4x - 24 = (x - 6)(x + 4)
\]
Set up the inequality:
\[
(x - 6)(x + 4) < 0
\]
Find the critical points:
\[
x = 6 \quad \text{and} \quad x = -4
\]
Test intervals: \( (-\infty, -4) \), \( (-4, 6) \), and \( (6, \infty) \).

- For \( x \in (-\infty, -4) \): Choose \( x = -5 \):
\[
(-5 - 6)(-5 + 4) = (-11)(-1) = 11 > 0
\]
- For \( x \in (-4, 6) \): Choose \( x = 0 \):
\[
(0 - 6)(0 + 4) = (-6)(4) = -24 < 0
\]
- For \( x \in (6, \infty) \): Choose \( x = 7 \):
\[
(7 - 6)(7 + 4) = (1)(11) = 11 > 0
\]

The inequality \( (x - 6)(x + 4) < 0 \) holds in the interval \( (-4, 6) \).

Solution:
\[
\boxed{(-4, 6)}
\]

---

#### 4. Solve \( x^2 + 8x \geq 0 \)

Factorize:
\[
x(x + 8) \geq 0
\]
Find the critical points:
\[
x = 0 \quad \text{and} \quad x = -8
\]
Test intervals: \( (-\infty, -8) \), \( (-8, 0) \), and \( (0, \infty) \).

- For \( x \in (-\infty, -8) \): Choose \( x = -9 \):
\[
(-9)(-9 + 8) = (-9)(-1) = 9 > 0
\]
- For \( x \in (-8, 0) \): Choose \( x = -1 \):
\[
(-1)(-1 + 8) = (-1)(7) = -7 < 0
\]
- For \( x \in (0, \infty) \): Choose \( x = 1 \):
\[
(1)(1 + 8) = (1)(9) = 9 > 0
\]

The inequality \( x(x + 8) \geq 0 \) holds in the intervals \( (-\infty, -8] \) and \( [0, \infty) \).

Solution:
\[
\boxed{(-\infty, -8] \cup [0, \infty)}
\]

---

#### 5. Solve \( x^2 - 6x - 16 \geq 0 \)

Factorize:
\[
x^2 - 6x - 16 = (x - 8)(x + 2)
\]
Set up the inequality:
\[
(x - 8)(x + 2) \geq 0
\]
Find the critical points:
\[
x = 8 \quad \text{and} \quad x = -2
\]
Test intervals: \( (-\infty, -2) \), \( (-2, 8) \), and \( (8, \infty) \).

- For \( x \in (-\infty, -2) \): Choose \( x = -3 \):
\[
(-3 - 8)(-3 + 2) = (-11)(-1) = 11 > 0
\]
- For \( x \in (-2, 8) \): Choose \( x = 0 \):
\[
(0 - 8)(0 + 2) = (-8)(2) = -16 < 0
\]
- For \( x \in (8, \infty) \): Choose \( x = 9 \):
\[
(9 - 8)(9 + 2) = (1)(11) = 11 > 0
\]

The inequality \( (x - 8)(x + 2) \geq 0 \) holds in the intervals \( (-\infty, -2] \) and \( [8, \infty) \).

Solution:
\[
\boxed{(-\infty, -2] \cup [8, \infty)}
\]

---

#### 6. Solve \( x^2 - 14x > 0 \)

Factorize:
\[
x(x - 14) > 0
\]
Find the critical points:
\[
x = 0 \quad \text{and} \quad x = 14
\]
Test intervals: \( (-\infty, 0) \), \( (0, 14) \), and \( (14, \infty) \).

- For \( x \in (-\infty, 0) \): Choose \( x = -1 \):
\[
(-1)(-1 - 14) = (-1)(-15) = 15 > 0
\]
- For \( x \in (0, 14) \): Choose \( x = 7 \):
\[
(7)(7 - 14) = (7)(-7) = -49 < 0
\]
- For \( x \in (14, \infty) \): Choose \( x = 15 \):
\[
(15)(15 - 14) = (15)(1) = 15 > 0
\]

The inequality \( x(x - 14) > 0 \) holds in the intervals \( (-\infty, 0) \) and \( (14, \infty) \).

Solution:
\[
\boxed{(-\infty, 0) \cup (14, \infty)}
\]

---

#### 7. Solve \( x^2 - 21x + 108 \leq 0 \)

Factorize:
\[
x^2 - 21x + 108 = (x - 9)(x - 12)
\]
Set up the inequality:
\[
(x - 9)(x - 12) \leq 0
\]
Find the critical points:
\[
x = 9 \quad \text{and} \quad x = 12
\]
Test intervals: \( (-\infty, 9) \), \( (9, 12) \), and \( (12, \infty) \).

- For \( x \in (-\infty, 9) \): Choose \( x = 8 \):
\[
(8 - 9)(8 - 12) = (-1)(-4) = 4 > 0
\]
- For \( x \in (9, 12) \): Choose \( x = 10 \):
\[
(10 - 9)(10 - 12) = (1)(-2) = -2 < 0
\]
- For \( x \in (12, \infty) \): Choose \( x = 13 \):
\[
(13 - 9)(13 - 12) = (4)(1) = 4 > 0
\]

The inequality \( (x - 9)(x - 12) \leq 0 \) holds in the interval \( [9, 12] \).

Solution:
\[
\boxed{[9, 12]}
\]

---

#### 8. Solve \( x^2 + 4x + 5 \geq 0 \)

First, solve the corresponding equation:
\[
x^2 + 4x + 5 = 0
\]
Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, b = 4, c = 5
\]
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}
\]
Since the discriminant (\( -4 \)) is negative, there are no real roots. The parabola \( x^2 + 4x + 5 \) opens upwards (since the coefficient of \( x^2 \) is positive), and it never touches the x-axis. Therefore, \( x^2 + 4x + 5 \) is always positive for all real \( x \).

Solution:
\[
\boxed{(-\infty, \infty)}
\]

---

#### 9. Solve \( y \leq x^2 - 8x + 16 \)

Rewrite the quadratic expression:
\[
x^2 - 8x + 16 = (x - 4)^2
\]
So the inequality becomes:
\[
y \leq (x - 4)^2
\]
This represents the region below or on the parabola \( y = (x - 4)^2 \).

---

#### 10. Solve \( y \geq x^2 + 4x - 6 \)

Rewrite the quadratic expression:
\[
x^2 + 4x - 6
\]
Complete the square:
\[
x^2 + 4x - 6 = (x + 2)^2 - 10
\]
So the inequality becomes:
\[
y \geq (x + 2)^2 - 10
\]
This represents the region above or on the parabola \( y = (x + 2)^2 - 10 \).

---

Final Answers:

1. \( \boxed{x = 0 \text{ or } x = 8} \)
2. \( \boxed{(-\infty, \infty)} \)
3. \( \boxed{(-4, 6)} \)
4. \( \boxed{(-\infty, -8] \cup [0, \infty)} \)
5. \( \boxed{(-\infty, -2] \cup [8, \infty)} \)
6. \( \boxed{(-\infty, 0) \cup (14, \infty)} \)
7. \( \boxed{[9, 12]} \)
8. \( \boxed{(-\infty, \infty)} \)
9. Region below or on \( y = (x - 4)^2 \)
10. Region above or on \( y = (x + 2)^2 - 10 \)

---

Boxed Final Answer:

\[
\boxed{
\begin{aligned}
1. & \quad x = 0 \text{ or } x = 8 \\
2. & \quad (-\infty, \infty) \\
3. & \quad (-4, 6) \\
4. & \quad (-\infty, -8] \cup [0, \infty) \\
5. & \quad (-\infty, -2] \cup [8, \infty) \\
6. & \quad (-\infty, 0) \cup (14, \infty) \\
7. & \quad [9, 12] \\
8. & \quad (-\infty, \infty) \\
9. & \quad y \leq (x - 4)^2 \\
10. & \quad y \geq (x + 2)^2 - 10
\end{aligned}
}
\]