Answer key for solving quadratic inequalities worksheet.

To solve quadratic inequalities, we follow a systematic approach that involves finding the roots of the corresponding quadratic equation, determining the intervals defined by these roots, and testing points in each interval to see where the inequality holds. Let's go through each problem step by step.

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Problem 1: \( 18x^2 + 23x + 5 \leq 0 \)

#### Step 1: Solve the quadratic equation \( 18x^2 + 23x + 5 = 0 \).
We use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 18 \), \( b = 23 \), and \( c = 5 \). Plugging in these values:
\[
x = \frac{-23 \pm \sqrt{23^2 - 4 \cdot 18 \cdot 5}}{2 \cdot 18} = \frac{-23 \pm \sqrt{529 - 360}}{36} = \frac{-23 \pm \sqrt{169}}{36} = \frac{-23 \pm 13}{36}
\]
This gives us two solutions:
\[
x = \frac{-23 + 13}{36} = \frac{-10}{36} = -\frac{5}{18}, \quad x = \frac{-23 - 13}{36} = \frac{-36}{36} = -1
\]
So, the roots are \( x = -1 \) and \( x = -\frac{5}{18} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, -\frac{5}{18}) \), and \( (-\frac{5}{18}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[
18(-2)^2 + 23(-2) + 5 = 72 - 46 + 5 = 31 > 0
\]
- For \( x \in (-1, -\frac{5}{18}) \), choose \( x = -\frac{3}{4} \):
\[
18\left(-\frac{3}{4}\right)^2 + 23\left(-\frac{3}{4}\right) + 5 = 18 \cdot \frac{9}{16} - \frac{69}{4} + 5 = \frac{162}{16} - \frac{276}{16} + \frac{80}{16} = \frac{-34}{16} < 0
\]
- For \( x \in (-\frac{5}{18}, \infty) \), choose \( x = 0 \):
\[
18(0)^2 + 23(0) + 5 = 5 > 0
\]

#### Step 4: Include the roots since the inequality is \( \leq 0 \).
The solution is:
\[
-1 \leq x \leq -\frac{5}{18}
\]

---

Problem 2: \( 12x^2 + 10x - 12 > 0 \)

#### Step 1: Solve the quadratic equation \( 12x^2 + 10x - 12 = 0 \).
Using the quadratic formula:
\[
x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 12 \cdot (-12)}}{2 \cdot 12} = \frac{-10 \pm \sqrt{100 + 576}}{24} = \frac{-10 \pm \sqrt{676}}{24} = \frac{-10 \pm 26}{24}
\]
This gives us two solutions:
\[
x = \frac{-10 + 26}{24} = \frac{16}{24} = \frac{2}{3}, \quad x = \frac{-10 - 26}{24} = \frac{-36}{24} = -\frac{3}{2}
\]
So, the roots are \( x = -\frac{3}{2} \) and \( x = \frac{2}{3} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, -\frac{3}{2}) \), \( (-\frac{3}{2}, \frac{2}{3}) \), and \( (\frac{2}{3}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, -\frac{3}{2}) \), choose \( x = -2 \):
\[
12(-2)^2 + 10(-2) - 12 = 48 - 20 - 12 = 16 > 0
\]
- For \( x \in (-\frac{3}{2}, \frac{2}{3}) \), choose \( x = 0 \):
\[
12(0)^2 + 10(0) - 12 = -12 < 0
\]
- For \( x \in (\frac{2}{3}, \infty) \), choose \( x = 1 \):
\[
12(1)^2 + 10(1) - 12 = 12 + 10 - 12 = 10 > 0
\]

#### Step 4: Exclude the roots since the inequality is \( > 0 \).
The solution is:
\[
x < -\frac{3}{2} \quad \text{or} \quad x > \frac{2}{3}
\]

---

Problem 3: \( -9x^2 + 29x - 6 \geq 0 \)

#### Step 1: Solve the quadratic equation \( -9x^2 + 29x - 6 = 0 \).
Using the quadratic formula:
\[
x = \frac{-29 \pm \sqrt{29^2 - 4 \cdot (-9) \cdot (-6)}}{2 \cdot (-9)} = \frac{-29 \pm \sqrt{841 - 216}}{-18} = \frac{-29 \pm \sqrt{625}}{-18} = \frac{-29 \pm 25}{-18}
\]
This gives us two solutions:
\[
x = \frac{-29 + 25}{-18} = \frac{-4}{-18} = \frac{2}{9}, \quad x = \frac{-29 - 25}{-18} = \frac{-54}{-18} = 3
\]
So, the roots are \( x = \frac{2}{9} \) and \( x = 3 \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, \frac{2}{9}) \), \( (\frac{2}{9}, 3) \), and \( (3, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, \frac{2}{9}) \), choose \( x = 0 \):
\[
-9(0)^2 + 29(0) - 6 = -6 < 0
\]
- For \( x \in (\frac{2}{9}, 3) \), choose \( x = 1 \):
\[
-9(1)^2 + 29(1) - 6 = -9 + 29 - 6 = 14 > 0
\]
- For \( x \in (3, \infty) \), choose \( x = 4 \):
\[
-9(4)^2 + 29(4) - 6 = -144 + 116 - 6 = -34 < 0
\]

#### Step 4: Include the roots since the inequality is \( \geq 0 \).
The solution is:
\[
\frac{2}{9} \leq x \leq 3
\]

---

Problem 4: \( 4x^2 + 20x - 11 < 0 \)

#### Step 1: Solve the quadratic equation \( 4x^2 + 20x - 11 = 0 \).
Using the quadratic formula:
\[
x = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 4 \cdot (-11)}}{2 \cdot 4} = \frac{-20 \pm \sqrt{400 + 176}}{8} = \frac{-20 \pm \sqrt{576}}{8} = \frac{-20 \pm 24}{8}
\]
This gives us two solutions:
\[
x = \frac{-20 + 24}{8} = \frac{4}{8} = \frac{1}{2}, \quad x = \frac{-20 - 24}{8} = \frac{-44}{8} = -\frac{11}{2}
\]
So, the roots are \( x = -\frac{11}{2} \) and \( x = \frac{1}{2} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, -\frac{11}{2}) \), \( (-\frac{11}{2}, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, -\frac{11}{2}) \), choose \( x = -6 \):
\[
4(-6)^2 + 20(-6) - 11 = 144 - 120 - 11 = 13 > 0
\]
- For \( x \in (-\frac{11}{2}, \frac{1}{2}) \), choose \( x = 0 \):
\[
4(0)^2 + 20(0) - 11 = -11 < 0
\]
- For \( x \in (\frac{1}{2}, \infty) \), choose \( x = 1 \):
\[
4(1)^2 + 20(1) - 11 = 4 + 20 - 11 = 13 > 0
\]

#### Step 4: Exclude the roots since the inequality is \( < 0 \).
The solution is:
\[
-\frac{11}{2} < x < \frac{1}{2}
\]

---

Problem 5: \( 7x^2 + 11x + 4 > 0 \)

#### Step 1: Solve the quadratic equation \( 7x^2 + 11x + 4 = 0 \).
Using the quadratic formula:
\[
x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 7 \cdot 4}}{2 \cdot 7} = \frac{-11 \pm \sqrt{121 - 112}}{14} = \frac{-11 \pm \sqrt{9}}{14} = \frac{-11 \pm 3}{14}
\]
This gives us two solutions:
\[
x = \frac{-11 + 3}{14} = \frac{-8}{14} = -\frac{4}{7}, \quad x = \frac{-11 - 3}{14} = \frac{-14}{14} = -1
\]
So, the roots are \( x = -1 \) and \( x = -\frac{4}{7} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, -\frac{4}{7}) \), and \( (-\frac{4}{7}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[
7(-2)^2 + 11(-2) + 4 = 28 - 22 + 4 = 10 > 0
\]
- For \( x \in (-1, -\frac{4}{7}) \), choose \( x = -\frac{3}{4} \):
\[
7\left(-\frac{3}{4}\right)^2 + 11\left(-\frac{3}{4}\right) + 4 = 7 \cdot \frac{9}{16} - \frac{33}{4} + 4 = \frac{63}{16} - \frac{132}{16} + \frac{64}{16} = \frac{-5}{16} < 0
\]
- For \( x \in (-\frac{4}{7}, \infty) \), choose \( x = 0 \):
\[
7(0)^2 + 11(0) + 4 = 4 > 0
\]

#### Step 4: Exclude the roots since the inequality is \( > 0 \).
The solution is:
\[
x < -1 \quad \text{or} \quad x > -\frac{4}{7}
\]

---

Problem 6: \( 17x^2 + 15x - 2 \geq 0 \)

#### Step 1: Solve the quadratic equation \( 17x^2 + 15x - 2 = 0 \).
Using the quadratic formula:
\[
x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 17 \cdot (-2)}}{2 \cdot 17} = \frac{-15 \pm \sqrt{225 + 136}}{34} = \frac{-15 \pm \sqrt{361}}{34} = \frac{-15 \pm 19}{34}
\]
This gives us two solutions:
\[
x = \frac{-15 + 19}{34} = \frac{4}{34} = \frac{2}{17}, \quad x = \frac{-15 - 19}{34} = \frac{-34}{34} = -1
\]
So, the roots are \( x = -1 \) and \( x = \frac{2}{17} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, \frac{2}{17}) \), and \( (\frac{2}{17}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[
17(-2)^2 + 15(-2) - 2 = 68 - 30 - 2 = 36 > 0
\]
- For \( x \in (-1, \frac{2}{17}) \), choose \( x = 0 \):
\[
17(0)^2 + 15(0) - 2 = -2 < 0
\]
- For \( x \in (\frac{2}{17}, \infty) \), choose \( x = 1 \):
\[
17(1)^2 + 15(1) - 2 = 17 + 15 - 2 = 30 > 0
\]

#### Step 4: Include the roots since the inequality is \( \geq 0 \).
The solution is:
\[
x \leq -1 \quad \text{or} \quad x \geq \frac{2}{17}
\]

---

Problem 7: \( 5x^2 - 15x + 10 < 0 \)

#### Step 1: Solve the quadratic equation \( 5x^2 - 15x + 10 = 0 \).
Factorize the quadratic:
\[
5(x^2 - 3x + 2) = 5(x - 1)(x - 2) = 0
\]
So, the roots are \( x = 1 \) and \( x = 2 \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, 1) \), choose \( x = 0 \):
\[
5(0)^2 - 15(0) + 10 = 10 > 0
\]
- For \( x \in (1, 2) \), choose \( x = 1.5 \):
\[
5(1.5)^2 - 15(1.5) + 10 = 5 \cdot 2.25 - 22.5 + 10 = 11.25 - 22.5 + 10 = -1.25 < 0
\]
- For \( x \in (2, \infty) \), choose \( x = 3 \):
\[
5(3)^2 - 15(3) + 10 = 45 - 45 + 10 = 10 > 0
\]

#### Step 4: Exclude the roots since the inequality is \( < 0 \).
The solution is:
\[
1 < x < 2
\]

---

Problem 8: \( -8x^2 + 6x - 1 \leq 0 \)

#### Step 1: Solve the quadratic equation \( -8x^2 + 6x - 1 = 0 \).
Using the quadratic formula:
\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot (-8) \cdot (-1)}}{2 \cdot (-8)} = \frac{-6 \pm \sqrt{36 - 32}}{-16} = \frac{-6 \pm \sqrt{4}}{-16} = \frac{-6 \pm 2}{-16}
\]
This gives us two solutions:
\[
x = \frac{-6 + 2}{-16} = \frac{-4}{-16} = \frac{1}{4}, \quad x = \frac{-6 - 2}{-16} = \frac{-8}{-16} = \frac{1}{2}
\]
So, the roots are \( x = \frac{1}{4} \) and \( x = \frac{1}{2} \).

#### Step 2: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, \frac{1}{4}) \), \( (\frac{1}{4}, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).

#### Step 3: Test points in each interval.
- For \( x \in (-\infty, \frac{1}{4}) \), choose \( x = 0 \):
\[
-8(0)^2 + 6(0) - 1 = -1 < 0
\]
- For \( x \in (\frac{1}{4}, \frac{1}{2}) \), choose \( x = \frac{1}{3} \):
\[
-8\left(\frac{1}{3}\right)^2 + 6\left(\frac{1}{3}\right) - 1 = -8 \cdot \frac{1}{9} + 2 - 1 = -\frac{8}{9} + 1 = \frac{1}{9} > 0
\]
- For \( x \in (\frac{1}{2}, \infty) \), choose \( x = 1 \):
\[
-8(1)^2 + 6(1) - 1 = -8 + 6 - 1 = -3 < 0
\]

#### Step 4: Include the roots since the inequality is \( \leq 0 \).
The solution is:
\[
x \leq \frac{1}{4} \quad \text{or} \quad x \geq \frac{1}{2}
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1) & \quad -1 \leq x \leq -\frac{5}{18} \\
2) & \quad x < -\frac{3}{2} \quad \text{or} \quad x > \frac{2}{3} \\
3) & \quad \frac{2}{9} \leq x \leq 3 \\
4) & \quad -\frac{11}{2} < x < \frac{1}{2} \\
5) & \quad x < -1 \quad \text{or} \quad x > -\frac{4}{7} \\
6) & \quad x \leq -1 \quad \text{or} \quad x \geq \frac{2}{17} \\
7) & \quad 1 < x < 2 \\
8) & \quad x \leq \frac{1}{4} \quad \text{or} \quad x \geq \frac{1}{2}
\end{aligned}
}
\]