Printable quadratic inequalities worksheet [PDFs] Brighterly.com - Free Printable

The task involves solving quadratic inequalities. To solve each inequality, we will follow these general steps:

1. Rewrite the inequality in standard form: Move all terms to one side of the inequality so that the other side is zero.
2. Factor the quadratic expression (if possible).
3. Find the critical points: Solve the equation obtained by setting the quadratic expression equal to zero.
4. Determine the intervals: Use the critical points to divide the real number line into intervals.
5. Test each interval: Choose a test point from each interval and substitute it into the original inequality to determine where the inequality holds true.
6. Write the solution: Express the solution in interval notation.

Let's solve each problem step by step.

---

Problem 1: \( x^2 + 6x - 6 < 1 \)

#### Step 1: Rewrite in standard form
\[ x^2 + 6x - 6 < 1 \]
\[ x^2 + 6x - 7 < 0 \]

#### Step 2: Factor the quadratic expression
\[ x^2 + 6x - 7 = (x + 7)(x - 1) \]

#### Step 3: Find the critical points
Set the factored expression equal to zero:
\[ (x + 7)(x - 1) = 0 \]
\[ x = -7 \quad \text{or} \quad x = 1 \]

#### Step 4: Determine the intervals
The critical points divide the real number line into three intervals:
\[ (-\infty, -7), \quad (-7, 1), \quad (1, \infty) \]

#### Step 5: Test each interval
- Interval \( (-\infty, -7) \): Choose \( x = -8 \)
\[ (-8 + 7)(-8 - 1) = (-1)(-9) = 9 \quad (\text{positive}) \]
This interval does not satisfy \( x^2 + 6x - 7 < 0 \).

- Interval \( (-7, 1) \): Choose \( x = 0 \)
\[ (0 + 7)(0 - 1) = (7)(-1) = -7 \quad (\text{negative}) \]
This interval satisfies \( x^2 + 6x - 7 < 0 \).

- Interval \( (1, \infty) \): Choose \( x = 2 \)
\[ (2 + 7)(2 - 1) = (9)(1) = 9 \quad (\text{positive}) \]
This interval does not satisfy \( x^2 + 6x - 7 < 0 \).

#### Step 6: Write the solution
The solution is:
\[ \boxed{(-7, 1)} \]

---

Problem 2: \( -4x^2 - 3 > -7x^2 + 5x - 5 \)

#### Step 1: Rewrite in standard form
\[ -4x^2 - 3 > -7x^2 + 5x - 5 \]
\[ -4x^2 + 7x^2 - 5x - 3 + 5 > 0 \]
\[ 3x^2 - 5x + 2 > 0 \]

#### Step 2: Factor the quadratic expression
\[ 3x^2 - 5x + 2 = (3x - 2)(x - 1) \]

#### Step 3: Find the critical points
Set the factored expression equal to zero:
\[ (3x - 2)(x - 1) = 0 \]
\[ x = \frac{2}{3} \quad \text{or} \quad x = 1 \]

#### Step 4: Determine the intervals
The critical points divide the real number line into three intervals:
\[ (-\infty, \frac{2}{3}), \quad \left( \frac{2}{3}, 1 \right), \quad (1, \infty) \]

#### Step 5: Test each interval
- Interval \( (-\infty, \frac{2}{3}) \): Choose \( x = 0 \)
\[ (3(0) - 2)(0 - 1) = (-2)(-1) = 2 \quad (\text{positive}) \]
This interval satisfies \( 3x^2 - 5x + 2 > 0 \).

- Interval \( \left( \frac{2}{3}, 1 \right) \): Choose \( x = \frac{3}{4} \)
\[ \left( 3\left(\frac{3}{4}\right) - 2 \right)\left(\frac{3}{4} - 1\right) = \left(\frac{9}{4} - 2\right)\left(\frac{3}{4} - 1\right) = \left(\frac{1}{4}\right)\left(-\frac{1}{4}\right) = -\frac{1}{16} \quad (\text{negative}) \]
This interval does not satisfy \( 3x^2 - 5x + 2 > 0 \).

- Interval \( (1, \infty) \): Choose \( x = 2 \)
\[ (3(2) - 2)(2 - 1) = (6 - 2)(1) = 4 \quad (\text{positive}) \]
This interval satisfies \( 3x^2 - 5x + 2 > 0 \).

#### Step 6: Write the solution
The solution is:
\[ \boxed{(-\infty, \frac{2}{3}) \cup (1, \infty)} \]

---

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

#### Step 1: Rewrite in standard form
\[ 2x^2 \geq 3x^2 + x - 4 \]
\[ 2x^2 - 3x^2 - x + 4 \geq 0 \]
\[ -x^2 - x + 4 \geq 0 \]
\[ x^2 + x - 4 \leq 0 \]

#### Step 2: Factor the quadratic expression
The quadratic \( x^2 + x - 4 \) does not factor nicely, so we use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), \( c = -4 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} \]
\[ x = \frac{-1 \pm \sqrt{1 + 16}}{2} \]
\[ x = \frac{-1 \pm \sqrt{17}}{2} \]

#### Step 3: Find the critical points
\[ x = \frac{-1 + \sqrt{17}}{2} \quad \text{or} \quad x = \frac{-1 - \sqrt{17}}{2} \]

#### Step 4: Determine the intervals
The critical points divide the real number line into three intervals:
\[ \left( -\infty, \frac{-1 - \sqrt{17}}{2} \right), \quad \left( \frac{-1 - \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2} \right), \quad \left( \frac{-1 + \sqrt{17}}{2}, \infty \right) \]

#### Step 5: Test each interval
- Interval \( \left( -\infty, \frac{-1 - \sqrt{17}}{2} \right) \): Choose a test point (e.g., a very small negative number).
- Interval \( \left( \frac{-1 - \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2} \right) \): Choose a test point (e.g., the midpoint).
- Interval \( \left( \frac{-1 + \sqrt{17}}{2}, \infty \right) \): Choose a test point (e.g., a very large positive number).

After testing, the middle interval satisfies \( x^2 + x - 4 \leq 0 \).

#### Step 6: Write the solution
The solution is:
\[ \boxed{\left[ \frac{-1 - \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2} \right]} \]

---

Final Answer

\[ \boxed{(-7, 1), \left( -\infty, \frac{2}{3} \right) \cup (1, \infty), \left[ \frac{-1 - \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2} \right]} \]