Solving Quadratic Inequalities Algebraically2 PDF | PDF | Algebra ... - Free Printable

To solve quadratic inequalities algebraically, we follow a systematic approach:

1. Rewrite the inequality in standard form (if necessary).
2. Find the roots of the corresponding quadratic equation by setting the inequality to zero.
3. Determine the intervals defined by the roots.
4. Test points in each interval to determine where the inequality holds true.
5. Consider the boundary points based on whether the inequality is strict (< or >) or non-strict (≤ or ≥).

Let's solve each problem step by step.

---

Problem 1: \( x^2 + x - 20 \leq 0 \)

#### Step 1: Solve the corresponding quadratic equation
\[ x^2 + x - 20 = 0 \]

Factorize:
\[ x^2 + x - 20 = (x + 5)(x - 4) = 0 \]

So, the roots are:
\[ x = -5 \quad \text{and} \quad x = 4 \]

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -5) \), choose \( x = -6 \):
\[ (-6)^2 + (-6) - 20 = 36 - 6 - 20 = 10 \quad (\text{positive}) \]

- For \( x \in (-5, 4) \), choose \( x = 0 \):
\[ 0^2 + 0 - 20 = -20 \quad (\text{negative}) \]

- For \( x \in (4, \infty) \), choose \( x = 5 \):
\[ 5^2 + 5 - 20 = 25 + 5 - 20 = 10 \quad (\text{positive}) \]

#### Step 4: Consider the inequality \( \leq 0 \)
The inequality \( x^2 + x - 20 \leq 0 \) holds where the expression is negative or zero. From the test points, this occurs in the interval \( (-5, 4) \). Since the inequality includes \( \leq \), the roots \( x = -5 \) and \( x = 4 \) are also included.

#### Solution:
\[ \boxed{[-5, 4]} \]

---

Problem 2: \( x^2 - 3x - 54 < 0 \)

#### Step 1: Solve the corresponding quadratic equation
\[ x^2 - 3x - 54 = 0 \]

Factorize:
\[ x^2 - 3x - 54 = (x - 9)(x + 6) = 0 \]

So, the roots are:
\[ x = 9 \quad \text{and} \quad x = -6 \]

#### Step 2: Determine the intervals
The roots divide the number line into three intervals:
\[ (-\infty, -6), \quad (-6, 9), \quad (9, \infty) \]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -6) \), choose \( x = -7 \):
\[ (-7)^2 - 3(-7) - 54 = 49 + 21 - 54 = 16 \quad (\text{positive}) \]

- For \( x \in (-6, 9) \), choose \( x = 0 \):
\[ 0^2 - 3(0) - 54 = -54 \quad (\text{negative}) \]

- For \( x \in (9, \infty) \), choose \( x = 10 \):
\[ 10^2 - 3(10) - 54 = 100 - 30 - 54 = 16 \quad (\text{positive}) \]

#### Step 4: Consider the inequality \( < 0 \)
The inequality \( x^2 - 3x - 54 < 0 \) holds where the expression is negative. From the test points, this occurs in the interval \( (-6, 9) \). Since the inequality is strict (<), the roots \( x = -6 \) and \( x = 9 \) are not included.

#### Solution:
\[ \boxed{(-6, 9)} \]

---

Problem 3: \( x^2 - 5x - 14 > 0 \)

#### Step 1: Solve the corresponding quadratic equation
\[ x^2 - 5x - 14 = 0 \]

Factorize:
\[ x^2 - 5x - 14 = (x - 7)(x + 2) = 0 \]

So, the roots are:
\[ x = 7 \quad \text{and} \quad x = -2 \]

#### Step 2: Determine the intervals
The roots divide the number line into three intervals:
\[ (-\infty, -2), \quad (-2, 7), \quad (7, \infty) \]

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
\[ (-3)^2 - 5(-3) - 14 = 9 + 15 - 14 = 10 \quad (\text{positive}) \]

- For \( x \in (-2, 7) \), choose \( x = 0 \):
\[ 0^2 - 5(0) - 14 = -14 \quad (\text{negative}) \]

- For \( x \in (7, \infty) \), choose \( x = 8 \):
\[ 8^2 - 5(8) - 14 = 64 - 40 - 14 = 10 \quad (\text{positive}) \]

#### Step 4: Consider the inequality \( > 0 \)
The inequality \( x^2 - 5x - 14 > 0 \) holds where the expression is positive. From the test points, this occurs in the intervals \( (-\infty, -2) \) and \( (7, \infty) \). Since the inequality is strict (>), the roots \( x = -2 \) and \( x = 7 \) are not included.

#### Solution:
\[ \boxed{(-\infty, -2) \cup (7, \infty)} \]

---

Problem 4: \( 2x^2 - 4x - 30 \geq 0 \)

#### Step 1: Simplify the inequality
Divide the entire inequality by 2:
\[ x^2 - 2x - 15 \geq 0 \]

#### Step 2: Solve the corresponding quadratic equation
\[ x^2 - 2x - 15 = 0 \]

Factorize:
\[ x^2 - 2x - 15 = (x - 5)(x + 3) = 0 \]

So, the roots are:
\[ x = 5 \quad \text{and} \quad x = -3 \]

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

#### Step 4: Test points in each interval
- For \( x \in (-\infty, -3) \), choose \( x = -4 \):
\[ (-4)^2 - 2(-4) - 15 = 16 + 8 - 15 = 9 \quad (\text{positive}) \]

- For \( x \in (-3, 5) \), choose \( x = 0 \):
\[ 0^2 - 2(0) - 15 = -15 \quad (\text{negative}) \]

- For \( x \in (5, \infty) \), choose \( x = 6 \):
\[ 6^2 - 2(6) - 15 = 36 - 12 - 15 = 9 \quad (\text{positive}) \]

#### Step 5: Consider the inequality \( \geq 0 \)
The inequality \( x^2 - 2x - 15 \geq 0 \) holds where the expression is positive or zero. From the test points, this occurs in the intervals \( (-\infty, -3] \) and \( [5, \infty) \). Since the inequality includes \( \geq \), the roots \( x = -3 \) and \( x = 5 \) are included.

#### Solution:
\[ \boxed{(-\infty, -3] \cup [5, \infty)} \]

---

Problem 5: \( 3x^2 - 6x - 9 \geq 0 \)

#### Step 1: Simplify the inequality
Factor out the common factor 3:
\[ 3(x^2 - 2x - 3) \geq 0 \]

Solve the corresponding quadratic equation:
\[ x^2 - 2x - 3 = 0 \]

Factorize:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \]

So, the roots are:
\[ x = 3 \quad \text{and} \quad x = -1 \]

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[ 3((-2)^2 - 2(-2) - 3) = 3(4 + 4 - 3) = 3(5) = 15 \quad (\text{positive}) \]

- For \( x \in (-1, 3) \), choose \( x = 0 \):
\[ 3(0^2 - 2(0) - 3) = 3(-3) = -9 \quad (\text{negative}) \]

- For \( x \in (3, \infty) \), choose \( x = 4 \):
\[ 3(4^2 - 2(4) - 3) = 3(16 - 8 - 3) = 3(5) = 15 \quad (\text{positive}) \]

#### Step 4: Consider the inequality \( \geq 0 \)
The inequality \( 3x^2 - 6x - 9 \geq 0 \) holds where the expression is positive or zero. From the test points, this occurs in the intervals \( (-\infty, -1] \) and \( [3, \infty) \). Since the inequality includes \( \geq \), the roots \( x = -1 \) and \( x = 3 \) are included.

#### Solution:
\[ \boxed{(-\infty, -1] \cup [3, \infty)} \]

---

Problem 6: \( 2x^2 + 9x - 5 < 0 \)

#### Step 1: Solve the corresponding quadratic equation
\[ 2x^2 + 9x - 5 = 0 \]

Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = 9 \), and \( c = -5 \).

\[ x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-5)}}{2(2)} \]
\[ x = \frac{-9 \pm \sqrt{81 + 40}}{4} \]
\[ x = \frac{-9 \pm \sqrt{121}}{4} \]
\[ x = \frac{-9 \pm 11}{4} \]

So, the roots are:
\[ x = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2} \]
\[ x = \frac{-9 - 11}{4} = \frac{-20}{4} = -5 \]

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

#### Step 3: Test points in each interval
- For \( x \in (-\infty, -5) \), choose \( x = -6 \):
\[ 2(-6)^2 + 9(-6) - 5 = 2(36) - 54 - 5 = 72 - 54 - 5 = 13 \quad (\text{positive}) \]

- For \( x \in (-5, \frac{1}{2}) \), choose \( x = 0 \):
\[ 2(0)^2 + 9(0) - 5 = -5 \quad (\text{negative}) \]

- For \( x \in (\frac{1}{2}, \infty) \), choose \( x = 1 \):
\[ 2(1)^2 + 9(1) - 5 = 2 + 9 - 5 = 6 \quad (\text{positive}) \]

#### Step 4: Consider the inequality \( < 0 \)
The inequality \( 2x^2 + 9x - 5 < 0 \) holds where the expression is negative. From the test points, this occurs in the interval \( (-5, \frac{1}{2}) \). Since the inequality is strict (<), the roots \( x = -5 \) and \( x = \frac{1}{2} \) are not included.

#### Solution:
\[ \boxed{(-5, \frac{1}{2})} \]

---

Final Answers:

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