Worksheet for solving quadratic inequalities.

To solve quadratic inequalities, we follow a systematic approach. Here's the general method:

Steps to Solve Quadratic Inequalities:

1. Rewrite the inequality in standard form: Ensure the inequality is in the form \( ax^2 + bx + c > 0 \) (or \( \geq 0 \), \( < 0 \), or \( \leq 0 \)).
2. Find the roots of the corresponding quadratic equation: Solve \( ax^2 + bx + c = 0 \) using factoring, completing the square, or the quadratic formula.
3. Determine the critical points: The roots of the quadratic equation are the critical points that divide the number line into intervals.
4. Test each interval: Choose a test point from each interval created by the critical points and substitute it into the original inequality to determine where the inequality holds true.
5. Consider the inequality sign: If the inequality is strict (\( > \) or \( < \)), exclude the critical points. If it is non-strict (\( \geq \) or \( \leq \)), include the critical points.
6. Write the solution in interval notation.

Now, let's solve each problem step by step.

---

Problem 1: \( -x^2 - 5x + 6 > 0 \)

#### Step 1: Rewrite the inequality
\[ -x^2 - 5x + 6 > 0 \]

#### Step 2: Find the roots
First, solve the equation \( -x^2 - 5x + 6 = 0 \). Factorize:
\[ -x^2 - 5x + 6 = -(x^2 + 5x - 6) = -(x + 6)(x - 1) \]
So, the roots are:
\[ x = -6 \quad \text{and} \quad x = 1 \]

#### Step 3: Determine the critical points
The critical points are \( x = -6 \) and \( x = 1 \). These divide the number line into three intervals: \( (-\infty, -6) \), \( (-6, 1) \), and \( (1, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -6) \), choose \( x = -7 \):
\[ -(x + 6)(x - 1) = -(-7 + 6)(-7 - 1) = -(-1)(-8) = -8 \quad (\text{not } > 0) \]
- For \( x \in (-6, 1) \), choose \( x = 0 \):
\[ -(x + 6)(x - 1) = -(0 + 6)(0 - 1) = -(6)(-1) = 6 \quad (> 0) \]
- For \( x \in (1, \infty) \), choose \( x = 2 \):
\[ -(x + 6)(x - 1) = -(2 + 6)(2 - 1) = -(8)(1) = -8 \quad (\text{not } > 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is strict (\( > 0 \)), the solution is the interval where the expression is positive:
\[ (-6, 1) \]

#### Final Answer:
\[ \boxed{(-6, 1)} \]

---

Problem 2: \( -x^2 - 12x - 11 \leq 0 \)

#### Step 1: Rewrite the inequality
\[ -x^2 - 12x - 11 \leq 0 \]

#### Step 2: Find the roots
Solve \( -x^2 - 12x - 11 = 0 \). Factorize:
\[ -x^2 - 12x - 11 = -(x^2 + 12x + 11) = -(x + 1)(x + 11) \]
So, the roots are:
\[ x = -1 \quad \text{and} \quad x = -11 \]

#### Step 3: Determine the critical points
The critical points are \( x = -11 \) and \( x = -1 \). These divide the number line into three intervals: \( (-\infty, -11) \), \( (-11, -1) \), and \( (-1, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -11) \), choose \( x = -12 \):
\[ -(x + 1)(x + 11) = -(-12 + 1)(-12 + 11) = -(-11)(-1) = -11 \quad (\text{not } \leq 0) \]
- For \( x \in (-11, -1) \), choose \( x = -6 \):
\[ -(x + 1)(x + 11) = -(-6 + 1)(-6 + 11) = -(-5)(5) = -25 \quad (\text{not } \leq 0) \]
- For \( x \in (-1, \infty) \), choose \( x = 0 \):
\[ -(x + 1)(x + 11) = -(0 + 1)(0 + 11) = -(1)(11) = -11 \quad (\text{not } \leq 0) \]

#### Step 5: Consider the inequality sign
The expression is zero at the critical points \( x = -11 \) and \( x = -1 \). Since the inequality is non-strict (\( \leq 0 \)), the solution includes these points:
\[ (-\infty, -11] \cup [-1, \infty) \]

#### Final Answer:
\[ \boxed{(-\infty, -11] \cup [-1, \infty)} \]

---

Problem 3: \( x^2 - 1 < 0 \)

#### Step 1: Rewrite the inequality
\[ x^2 - 1 < 0 \]

#### Step 2: Find the roots
Solve \( x^2 - 1 = 0 \). Factorize:
\[ x^2 - 1 = (x - 1)(x + 1) \]
So, the roots are:
\[ x = 1 \quad \text{and} \quad x = -1 \]

#### Step 3: Determine the critical points
The critical points are \( x = -1 \) and \( x = 1 \). These divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[ (x - 1)(x + 1) = (-2 - 1)(-2 + 1) = (-3)(-1) = 3 \quad (\text{not } < 0) \]
- For \( x \in (-1, 1) \), choose \( x = 0 \):
\[ (x - 1)(x + 1) = (0 - 1)(0 + 1) = (-1)(1) = -1 \quad (< 0) \]
- For \( x \in (1, \infty) \), choose \( x = 2 \):
\[ (x - 1)(x + 1) = (2 - 1)(2 + 1) = (1)(3) = 3 \quad (\text{not } < 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is strict (\( < 0 \)), the solution is the interval where the expression is negative:
\[ (-1, 1) \]

#### Final Answer:
\[ \boxed{(-1, 1)} \]

---

Problem 4: \( x^2 - 2x - 3 \geq 0 \)

#### Step 1: Rewrite the inequality
\[ x^2 - 2x - 3 \geq 0 \]

#### Step 2: Find the roots
Solve \( x^2 - 2x - 3 = 0 \). Factorize:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]
So, the roots are:
\[ x = 3 \quad \text{and} \quad x = -1 \]

#### Step 3: Determine the critical points
The critical points are \( x = -1 \) and \( x = 3 \). These divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[ (x - 3)(x + 1) = (-2 - 3)(-2 + 1) = (-5)(-1) = 5 \quad (\geq 0) \]
- For \( x \in (-1, 3) \), choose \( x = 0 \):
\[ (x - 3)(x + 1) = (0 - 3)(0 + 1) = (-3)(1) = -3 \quad (\text{not } \geq 0) \]
- For \( x \in (3, \infty) \), choose \( x = 4 \):
\[ (x - 3)(x + 1) = (4 - 3)(4 + 1) = (1)(5) = 5 \quad (\geq 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is non-strict (\( \geq 0 \)), the solution includes the critical points:
\[ (-\infty, -1] \cup [3, \infty) \]

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

---

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

#### Step 1: Rewrite the inequality
\[ x^2 + 4x + 5 > 0 \]

#### Step 2: Find the roots
Solve \( x^2 + 4x + 5 = 0 \). Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} \]
Since the discriminant is negative, there are no real roots. The quadratic expression \( x^2 + 4x + 5 \) is always positive for all real \( x \).

#### Final Answer:
\[ \boxed{(-\infty, \infty)} \]

---

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

#### Step 1: Rewrite the inequality
\[ x^2 - 5x - 6 < 0 \]

#### Step 2: Find the roots
Solve \( x^2 - 5x - 6 = 0 \). Factorize:
\[ x^2 - 5x - 6 = (x - 6)(x + 1) \]
So, the roots are:
\[ x = 6 \quad \text{and} \quad x = -1 \]

#### Step 3: Determine the critical points
The critical points are \( x = -1 \) and \( x = 6 \). These divide the number line into three intervals: \( (-\infty, -1) \), \( (-1, 6) \), and \( (6, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -1) \), choose \( x = -2 \):
\[ (x - 6)(x + 1) = (-2 - 6)(-2 + 1) = (-8)(-1) = 8 \quad (\text{not } < 0) \]
- For \( x \in (-1, 6) \), choose \( x = 0 \):
\[ (x - 6)(x + 1) = (0 - 6)(0 + 1) = (-6)(1) = -6 \quad (< 0) \]
- For \( x \in (6, \infty) \), choose \( x = 7 \):
\[ (x - 6)(x + 1) = (7 - 6)(7 + 1) = (1)(8) = 8 \quad (\text{not } < 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is strict (\( < 0 \)), the solution is the interval where the expression is negative:
\[ (-1, 6) \]

#### Final Answer:
\[ \boxed{(-1, 6)} \]

---

Problem 7: \( -x^2 + 3x + 10 \leq 0 \)

#### Step 1: Rewrite the inequality
\[ -x^2 + 3x + 10 \leq 0 \]

#### Step 2: Find the roots
Solve \( -x^2 + 3x + 10 = 0 \). Factorize:
\[ -x^2 + 3x + 10 = -(x^2 - 3x - 10) = -(x - 5)(x + 2) \]
So, the roots are:
\[ x = 5 \quad \text{and} \quad x = -2 \]

#### Step 3: Determine the critical points
The critical points are \( x = -2 \) and \( x = 5 \). These divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, 5) \), and \( (5, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -2) \), choose \( x = -3 \):
\[ -(x - 5)(x + 2) = -(-3 - 5)(-3 + 2) = -(-8)(-1) = -8 \quad (\text{not } \leq 0) \]
- For \( x \in (-2, 5) \), choose \( x = 0 \):
\[ -(x - 5)(x + 2) = -(0 - 5)(0 + 2) = -( -5)(2) = 10 \quad (\text{not } \leq 0) \]
- For \( x \in (5, \infty) \), choose \( x = 6 \):
\[ -(x - 5)(x + 2) = -(6 - 5)(6 + 2) = -(1)(8) = -8 \quad (\leq 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is non-strict (\( \leq 0 \)), the solution includes the critical points:
\[ (-\infty, -2] \cup [5, \infty) \]

#### Final Answer:
\[ \boxed{(-\infty, -2] \cup [5, \infty)} \]

---

Problem 8: \( x^2 + 8x - 9 \geq 0 \)

#### Step 1: Rewrite the inequality
\[ x^2 + 8x - 9 \geq 0 \]

#### Step 2: Find the roots
Solve \( x^2 + 8x - 9 = 0 \). Factorize:
\[ x^2 + 8x - 9 = (x + 9)(x - 1) \]
So, the roots are:
\[ x = -9 \quad \text{and} \quad x = 1 \]

#### Step 3: Determine the critical points
The critical points are \( x = -9 \) and \( x = 1 \). These divide the number line into three intervals: \( (-\infty, -9) \), \( (-9, 1) \), and \( (1, \infty) \).

#### Step 4: Test each interval
Choose test points from each interval:
- For \( x \in (-\infty, -9) \), choose \( x = -10 \):
\[ (x + 9)(x - 1) = (-10 + 9)(-10 - 1) = (-1)(-11) = 11 \quad (\geq 0) \]
- For \( x \in (-9, 1) \), choose \( x = 0 \):
\[ (x + 9)(x - 1) = (0 + 9)(0 - 1) = (9)(-1) = -9 \quad (\text{not } \geq 0) \]
- For \( x \in (1, \infty) \), choose \( x = 2 \):
\[ (x + 9)(x - 1) = (2 + 9)(2 - 1) = (11)(1) = 11 \quad (\geq 0) \]

#### Step 5: Consider the inequality sign
Since the inequality is non-strict (\( \geq 0 \)), the solution includes the critical points:
\[ (-\infty, -9] \cup [1, \infty) \]

#### Final Answer:
\[ \boxed{(-\infty, -9] \cup [1, \infty)} \]

---

Final Answers for All Problems:

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