A22c - Solving quadratic inequalities in one variable - BossMaths.com - Free Printable

To solve the given quadratic inequalities, we will follow a systematic approach for each problem. The general steps are:

1. Rewrite the inequality in standard form (if necessary).
2. Find the roots of the corresponding quadratic equation.
3. Determine the intervals defined by the roots.
4. Test points in each interval to determine where the inequality holds.
5. Write the solution in interval notation.

Let's solve each inequality step by step.

---

1) \( 2x^2 + 5x + 2 < 0 \)

#### Step 1: Find the roots of \( 2x^2 + 5x + 2 = 0 \)
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 2, \, b = 5, \, c = 2
\]
\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm \sqrt{9}}{4} = \frac{-5 \pm 3}{4}
\]
\[
x = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2}, \quad x = \frac{-5 - 3}{4} = \frac{-8}{4} = -2
\]
The roots are \( x = -2 \) and \( x = -\frac{1}{2} \).

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
\[
2(-3)^2 + 5(-3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 5 > 0
\]
- For \( x \in (-2, -\frac{1}{2}) \), choose \( x = -1 \):
\[
2(-1)^2 + 5(-1) + 2 = 2(1) - 5 + 2 = 2 - 5 + 2 = -1 < 0
\]
- For \( x \in (-\frac{1}{2}, \infty) \), choose \( x = 0 \):
\[
2(0)^2 + 5(0) + 2 = 2 > 0
\]

#### Step 4: Write the solution
The inequality \( 2x^2 + 5x + 2 < 0 \) holds in the interval \( (-2, -\frac{1}{2}) \).

\[
\boxed{(-2, -\frac{1}{2})}
\]

---

2) \( 2x^2 - 15x + 7 > 0 \)

#### Step 1: Find the roots of \( 2x^2 - 15x + 7 = 0 \)
Using the quadratic formula:
\[
a = 2, \, b = -15, \, c = 7
\]
\[
x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 7}}{2 \cdot 2} = \frac{15 \pm \sqrt{225 - 56}}{4} = \frac{15 \pm \sqrt{169}}{4} = \frac{15 \pm 13}{4}
\]
\[
x = \frac{15 + 13}{4} = \frac{28}{4} = 7, \quad x = \frac{15 - 13}{4} = \frac{2}{4} = \frac{1}{2}
\]
The roots are \( x = \frac{1}{2} \) and \( x = 7 \).

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, \frac{1}{2}) \), choose \( x = 0 \):
\[
2(0)^2 - 15(0) + 7 = 7 > 0
\]
- For \( x \in (\frac{1}{2}, 7) \), choose \( x = 1 \):
\[
2(1)^2 - 15(1) + 7 = 2 - 15 + 7 = -6 < 0
\]
- For \( x \in (7, \infty) \), choose \( x = 8 \):
\[
2(8)^2 - 15(8) + 7 = 2(64) - 120 + 7 = 128 - 120 + 7 = 15 > 0
\]

#### Step 4: Write the solution
The inequality \( 2x^2 - 15x + 7 > 0 \) holds in the intervals \( (-\infty, \frac{1}{2}) \) and \( (7, \infty) \).

\[
\boxed{(-\infty, \frac{1}{2}) \cup (7, \infty)}
\]

---

3) \( -3x^2 - 4x + 4 > 0 \)

#### Step 1: Find the roots of \( -3x^2 - 4x + 4 = 0 \)
Using the quadratic formula:
\[
a = -3, \, b = -4, \, c = 4
\]
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot (-3) \cdot 4}}{2 \cdot (-3)} = \frac{4 \pm \sqrt{16 + 48}}{-6} = \frac{4 \pm \sqrt{64}}{-6} = \frac{4 \pm 8}{-6}
\]
\[
x = \frac{4 + 8}{-6} = \frac{12}{-6} = -2, \quad x = \frac{4 - 8}{-6} = \frac{-4}{-6} = \frac{2}{3}
\]
The roots are \( x = -2 \) and \( x = \frac{2}{3} \).

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
\[
-3(-3)^2 - 4(-3) + 4 = -3(9) + 12 + 4 = -27 + 12 + 4 = -11 < 0
\]
- For \( x \in (-2, \frac{2}{3}) \), choose \( x = 0 \):
\[
-3(0)^2 - 4(0) + 4 = 4 > 0
\]
- For \( x \in (\frac{2}{3}, \infty) \), choose \( x = 1 \):
\[
-3(1)^2 - 4(1) + 4 = -3 - 4 + 4 = -3 < 0
\]

#### Step 4: Write the solution
The inequality \( -3x^2 - 4x + 4 > 0 \) holds in the interval \( (-2, \frac{2}{3}) \).

\[
\boxed{(-2, \frac{2}{3})}
\]

---

4) \( -x^2 - x \leq 0 \)

#### Step 1: Factor the expression
\[
-x^2 - x = -x(x + 1)
\]
The roots are \( x = 0 \) and \( x = -1 \).

#### Step 2: Determine the intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -1), \, (-1, 0), \, (0, \infty)
\]

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

#### Step 4: Include the roots
The inequality \( -x^2 - x \leq 0 \) holds at the roots \( x = -1 \) and \( x = 0 \).

#### Step 5: Write the solution
The inequality \( -x^2 - x \leq 0 \) holds in the intervals \( (-\infty, -1] \) and \( [0, \infty) \).

\[
\boxed{(-\infty, -1] \cup [0, \infty)}
\]

---

5) \( -x^2 - 8x - 15 \geq 0 \)

#### Step 1: Factor the expression
\[
-x^2 - 8x - 15 = -(x^2 + 8x + 15) = -(x + 3)(x + 5)
\]
The roots are \( x = -3 \) and \( x = -5 \).

#### Step 2: Determine the intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -5), \, (-5, -3), \, (-3, \infty)
\]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -5) \), choose \( x = -6 \):
\[
-(-6)^2 - 8(-6) - 15 = -36 + 48 - 15 = -3 < 0
\]
- For \( x \in (-5, -3) \), choose \( x = -4 \):
\[
-(-4)^2 - 8(-4) - 15 = -16 + 32 - 15 = 1 > 0
\]
- For \( x \in (-3, \infty) \), choose \( x = -2 \):
\[
-(-2)^2 - 8(-2) - 15 = -4 + 16 - 15 = -3 < 0
\]

#### Step 4: Include the roots
The inequality \( -x^2 - 8x - 15 \geq 0 \) holds at the roots \( x = -5 \) and \( x = -3 \).

#### Step 5: Write the solution
The inequality \( -x^2 - 8x - 15 \geq 0 \) holds in the interval \( [-5, -3] \).

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

---

6) \( 4x^2 - 4x - 15 < 0 \)

#### Step 1: Find the roots of \( 4x^2 - 4x - 15 = 0 \)
Using the quadratic formula:
\[
a = 4, \, b = -4, \, c = -15
\]
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-15)}}{2 \cdot 4} = \frac{4 \pm \sqrt{16 + 240}}{8} = \frac{4 \pm \sqrt{256}}{8} = \frac{4 \pm 16}{8}
\]
\[
x = \frac{4 + 16}{8} = \frac{20}{8} = \frac{5}{2}, \quad x = \frac{4 - 16}{8} = \frac{-12}{8} = -\frac{3}{2}
\]
The roots are \( x = -\frac{3}{2} \) and \( x = \frac{5}{2} \).

#### Step 2: Determine the intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -\frac{3}{2}), \, (-\frac{3}{2}, \frac{5}{2}), \, (\frac{5}{2}, \infty)
\]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -\frac{3}{2}) \), choose \( x = -2 \):
\[
4(-2)^2 - 4(-2) - 15 = 4(4) + 8 - 15 = 16 + 8 - 15 = 9 > 0
\]
- For \( x \in (-\frac{3}{2}, \frac{5}{2}) \), choose \( x = 0 \):
\[
4(0)^2 - 4(0) - 15 = -15 < 0
\]
- For \( x \in (\frac{5}{2}, \infty) \), choose \( x = 3 \):
\[
4(3)^2 - 4(3) - 15 = 4(9) - 12 - 15 = 36 - 12 - 15 = 9 > 0
\]

#### Step 4: Write the solution
The inequality \( 4x^2 - 4x - 15 < 0 \) holds in the interval \( (-\frac{3}{2}, \frac{5}{2}) \).

\[
\boxed{(-\frac{3}{2}, \frac{5}{2})}
\]

---

7) \( 6x^2 - x - 2 < 0 \)

#### Step 1: Find the roots of \( 6x^2 - x - 2 = 0 \)
Using the quadratic formula:
\[
a = 6, \, b = -1, \, c = -2
\]
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6} = \frac{1 \pm \sqrt{1 + 48}}{12} = \frac{1 \pm \sqrt{49}}{12} = \frac{1 \pm 7}{12}
\]
\[
x = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}, \quad x = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
The roots are \( x = -\frac{1}{2} \) and \( x = \frac{2}{3} \).

#### Step 2: Determine the intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -\frac{1}{2}), \, (-\frac{1}{2}, \frac{2}{3}), \, (\frac{2}{3}, \infty)
\]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -\frac{1}{2}) \), choose \( x = -1 \):
\[
6(-1)^2 - (-1) - 2 = 6 + 1 - 2 = 5 > 0
\]
- For \( x \in (-\frac{1}{2}, \frac{2}{3}) \), choose \( x = 0 \):
\[
6(0)^2 - (0) - 2 = -2 < 0
\]
- For \( x \in (\frac{2}{3}, \infty) \), choose \( x = 1 \):
\[
6(1)^2 - (1) - 2 = 6 - 1 - 2 = 3 > 0
\]

#### Step 4: Write the solution
The inequality \( 6x^2 - x - 2 < 0 \) holds in the interval \( (-\frac{1}{2}, \frac{2}{3}) \).

\[
\boxed{(-\frac{1}{2}, \frac{2}{3})}
\]

---

8) \( -4x^2 + 3x + 10 \geq 0 \)

#### Step 1: Find the roots of \( -4x^2 + 3x + 10 = 0 \)
Using the quadratic formula:
\[
a = -4, \, b = 3, \, c = 10
\]
\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot (-4) \cdot 10}}{2 \cdot (-4)} = \frac{-3 \pm \sqrt{9 + 160}}{-8} = \frac{-3 \pm \sqrt{169}}{-8} = \frac{-3 \pm 13}{-8}
\]
\[
x = \frac{-3 + 13}{-8} = \frac{10}{-8} = -\frac{5}{4}, \quad x = \frac{-3 - 13}{-8} = \frac{-16}{-8} = 2
\]
The roots are \( x = -\frac{5}{4} \) and \( x = 2 \).

#### Step 2: Determine the intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -\frac{5}{4}), \, (-\frac{5}{4}, 2), \, (2, \infty)
\]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -\frac{5}{4}) \), choose \( x = -2 \):
\[
-4(-2)^2 + 3(-2) + 10 = -4(4) - 6 + 10 = -16 - 6 + 10 = -12 < 0
\]
- For \( x \in (-\frac{5}{4}, 2) \), choose \( x = 0 \):
\[
-4(0)^2 + 3(0) + 10 = 10 > 0
\]
- For \( x \in (2, \infty) \), choose \( x = 3 \):
\[
-4(3)^2 + 3(3) + 10 = -4(9) + 9 + 10 = -36 + 9 + 10 = -17 < 0
\]

#### Step 4: Include the roots
The inequality \( -4x^2 + 3x + 10 \geq 0 \) holds at the roots \( x = -\frac{5}{4} \) and \( x = 2 \).

#### Step 5: Write the solution
The inequality \( -4x^2 + 3x + 10 \geq 0 \) holds in the interval \( \left[ -\frac{5}{4}, 2 \right] \).

\[
\boxed{\left[ -\frac{5}{4}, 2 \right]}
\]

---

Final Answers

\[
\boxed{
\begin{aligned}
1) & \, (-2, -\frac{1}{2}) \\
2) & \, (-\infty, \frac{1}{2}) \cup (7, \infty) \\
3) & \, (-2, \frac{2}{3}) \\
4) & \, (-\infty, -1] \cup [0, \infty) \\
5) & \, [-5, -3] \\
6) & \, (-\frac{3}{2}, \frac{5}{2}) \\
7) & \, (-\frac{1}{2}, \frac{2}{3}) \\
8) & \, \left[ -\frac{5}{4}, 2 \right]
\end{aligned}
}
\]