Grade mathematics: Quadratic Inequalities | PPT - Free Printable

To solve the given linear inequalities, we will address each one step by step. Let's go through them systematically.

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1. \( x^2 - 6x + 8 \geq 0 \)

#### Step 1: Factorize the quadratic expression.
\[ x^2 - 6x + 8 = (x - 4)(x - 2) \]

#### Step 2: Find the roots.
The roots are \( x = 4 \) and \( x = 2 \).

#### Step 3: Determine the intervals.
The roots divide the number line into three intervals: \( (-\infty, 2) \), \( (2, 4) \), and \( (4, \infty) \).

#### Step 4: Test points in each interval.
- For \( x \in (-\infty, 2) \): Choose \( x = 0 \).
\[ (0 - 4)(0 - 2) = (-4)(-2) = 8 > 0 \]
- For \( x \in (2, 4) \): Choose \( x = 3 \).
\[ (3 - 4)(3 - 2) = (-1)(1) = -1 < 0 \]
- For \( x \in (4, \infty) \): Choose \( x = 5 \).
\[ (5 - 4)(5 - 2) = (1)(3) = 3 > 0 \]

#### Step 5: Include the roots since the inequality is \( \geq 0 \).
The solution is:
\[ x \in (-\infty, 2] \cup [4, \infty) \]

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2. \( x^2 + 2x + 1 \geq 0 \)

#### Step 1: Factorize the quadratic expression.
\[ x^2 + 2x + 1 = (x + 1)^2 \]

#### Step 2: Analyze the expression.
Since \( (x + 1)^2 \) is a perfect square, it is always non-negative (\( \geq 0 \)) for all real \( x \).

#### Step 3: Determine when it equals zero.
\[ (x + 1)^2 = 0 \implies x = -1 \]

#### Step 4: Solution.
The solution is:
\[ x \in (-\infty, \infty) \]

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3. \( x^2 + x + 1 > 0 \)

#### Step 1: Analyze the quadratic expression.
The quadratic \( x^2 + x + 1 \) has no real roots because its discriminant is negative:
\[ \Delta = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 \]

#### Step 2: Determine the sign of the quadratic.
Since the leading coefficient is positive (\( a = 1 \)), the parabola opens upwards, and the quadratic is always positive.

#### Step 3: Solution.
The solution is:
\[ x \in (-\infty, \infty) \]

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4. \( 7x^2 + 2x - 28 < 0 \)

#### Step 1: Solve the corresponding equation.
\[ 7x^2 + 2x - 28 = 0 \]

Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 7 \), \( b = 2 \), and \( c = -28 \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4(7)(-28)}}{2(7)} \]
\[ x = \frac{-2 \pm \sqrt{4 + 784}}{14} \]
\[ x = \frac{-2 \pm \sqrt{788}}{14} \]
\[ x = \frac{-2 \pm 2\sqrt{197}}{14} \]
\[ x = \frac{-1 \pm \sqrt{197}}{7} \]

#### Step 2: Determine the intervals.
The roots are \( x = \frac{-1 + \sqrt{197}}{7} \) and \( x = \frac{-1 - \sqrt{197}}{7} \). These roots divide the number line into three intervals.

#### Step 3: Test points in each interval.
- For \( x \in \left( -\infty, \frac{-1 - \sqrt{197}}{7} \right) \): Choose a point and test.
- For \( x \in \left( \frac{-1 - \sqrt{197}}{7}, \frac{-1 + \sqrt{197}}{7} \right) \): Choose a point and test.
- For \( x \in \left( \frac{-1 + \sqrt{197}}{7}, \infty \right) \): Choose a point and test.

The quadratic is negative between the roots.

#### Step 4: Solution.
The solution is:
\[ x \in \left( \frac{-1 - \sqrt{197}}{7}, \frac{-1 + \sqrt{197}}{7} \right) \]

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5. \( -x^2 + 4x - 7 < 0 \)

#### Step 1: Rewrite the inequality.
\[ -(x^2 - 4x + 7) < 0 \]
\[ x^2 - 4x + 7 > 0 \]

#### Step 2: Analyze the quadratic expression.
The quadratic \( x^2 - 4x + 7 \) has no real roots because its discriminant is negative:
\[ \Delta = b^2 - 4ac = (-4)^2 - 4(1)(7) = 16 - 28 = -12 \]

#### Step 3: Determine the sign of the quadratic.
Since the leading coefficient is positive (\( a = 1 \)), the parabola opens upwards, and the quadratic is always positive.

#### Step 4: Solution.
The solution is:
\[ x \in (-\infty, \infty) \]

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6. \( 4x^2 - 16 > 0 \)

#### Step 1: Simplify the inequality.
\[ 4(x^2 - 4) > 0 \]
\[ x^2 - 4 > 0 \]
\[ (x - 2)(x + 2) > 0 \]

#### Step 2: Find the roots.
The roots are \( x = 2 \) and \( x = -2 \).

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

#### Step 4: Test points in each interval.
- For \( x \in (-\infty, -2) \): Choose \( x = -3 \).
\[ (-3 - 2)(-3 + 2) = (-5)(-1) = 5 > 0 \]
- For \( x \in (-2, 2) \): Choose \( x = 0 \).
\[ (0 - 2)(0 + 2) = (-2)(2) = -4 < 0 \]
- For \( x \in (2, \infty) \): Choose \( x = 3 \).
\[ (3 - 2)(3 + 2) = (1)(5) = 5 > 0 \]

#### Step 5: Exclude the roots since the inequality is \( > 0 \).
The solution is:
\[ x \in (-\infty, -2) \cup (2, \infty) \]

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7. \( 4x^2 - 4x + 1 \leq 0 \)

#### Step 1: Factorize the quadratic expression.
\[ 4x^2 - 4x + 1 = (2x - 1)^2 \]

#### Step 2: Analyze the expression.
Since \( (2x - 1)^2 \) is a perfect square, it is always non-negative (\( \geq 0 \)) for all real \( x \).

#### Step 3: Determine when it equals zero.
\[ (2x - 1)^2 = 0 \implies x = \frac{1}{2} \]

#### Step 4: Solution.
The solution is:
\[ x = \frac{1}{2} \]

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8. \( 8x^4 + 12x^2 - 64x^2 > 0 \)

#### Step 1: Simplify the inequality.
\[ 8x^4 + 12x^2 - 64x^2 > 0 \]
\[ 8x^4 - 52x^2 > 0 \]
\[ 4x^2(2x^2 - 13) > 0 \]

#### Step 2: Factorize.
\[ 4x^2(2x^2 - 13) > 0 \]

#### Step 3: Find the roots.
The roots are \( x = 0 \) and \( x = \pm \sqrt{\frac{13}{2}} \).

#### Step 4: Determine the intervals.
The roots divide the number line into intervals: \( (-\infty, -\sqrt{\frac{13}{2}}) \), \( (-\sqrt{\frac{13}{2}}, 0) \), \( (0, \sqrt{\frac{13}{2}}) \), and \( (\sqrt{\frac{13}{2}}, \infty) \).

#### Step 5: Test points in each interval.
- For \( x \in (-\infty, -\sqrt{\frac{13}{2}}) \): Choose a point and test.
- For \( x \in (-\sqrt{\frac{13}{2}}, 0) \): Choose a point and test.
- For \( x \in (0, \sqrt{\frac{13}{2}}) \): Choose a point and test.
- For \( x \in (\sqrt{\frac{13}{2}}, \infty) \): Choose a point and test.

The expression is positive in \( (-\infty, -\sqrt{\frac{13}{2}}) \) and \( (\sqrt{\frac{13}{2}}, \infty) \).

#### Step 6: Exclude the roots since the inequality is \( > 0 \).
The solution is:
\[ x \in (-\infty, -\sqrt{\frac{13}{2}}) \cup (\sqrt{\frac{13}{2}}, \infty) \]

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9. \( x^4 - 25x^2 + 144 < 0 \)

#### Step 1: Let \( y = x^2 \).
The inequality becomes:
\[ y^2 - 25y + 144 < 0 \]

#### Step 2: Solve the quadratic inequality.
Factorize:
\[ y^2 - 25y + 144 = (y - 9)(y - 16) \]

#### Step 3: Find the roots.
The roots are \( y = 9 \) and \( y = 16 \).

#### Step 4: Determine the intervals.
The roots divide the number line into intervals: \( (-\infty, 9) \), \( (9, 16) \), and \( (16, \infty) \).

#### Step 5: Test points in each interval.
- For \( y \in (9, 16) \): The expression is negative.

#### Step 6: Convert back to \( x \).
Since \( y = x^2 \), we have:
\[ 9 < x^2 < 16 \]
\[ 3 < |x| < 4 \]
\[ x \in (-4, -3) \cup (3, 4) \]

#### Step 7: Solution.
The solution is:
\[ x \in (-4, -3) \cup (3, 4) \]

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10. \( x^4 - 16x^2 - 225 \geq 0 \)

#### Step 1: Let \( y = x^2 \).
The inequality becomes:
\[ y^2 - 16y - 225 \geq 0 \]

#### Step 2: Solve the quadratic inequality.
Factorize:
\[ y^2 - 16y - 225 = (y - 25)(y + 9) \]

#### Step 3: Find the roots.
The roots are \( y = 25 \) and \( y = -9 \).

#### Step 4: Determine the intervals.
The roots divide the number line into intervals: \( (-\infty, -9) \), \( (-9, 25) \), and \( (25, \infty) \).

#### Step 5: Test points in each interval.
- For \( y \in (-\infty, -9) \): The expression is positive.
- For \( y \in (-9, 25) \): The expression is negative.
- For \( y \in (25, \infty) \): The expression is positive.

#### Step 6: Include the roots since the inequality is \( \geq 0 \).
The solution for \( y \) is:
\[ y \in (-\infty, -9] \cup [25, \infty) \]

#### Step 7: Convert back to \( x \).
Since \( y = x^2 \), we have:
\[ x^2 \geq 25 \]
\[ |x| \geq 5 \]
\[ x \in (-\infty, -5] \cup [5, \infty) \]

#### Step 8: Solution.
The solution is:
\[ x \in (-\infty, -5] \cup [5, \infty) \]

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Final Answers:

1. \( x \in (-\infty, 2] \cup [4, \infty) \)
2. \( x \in (-\infty, \infty) \)
3. \( x \in (-\infty, \infty) \)
4. \( x \in \left( \frac{-1 - \sqrt{197}}{7}, \frac{-1 + \sqrt{197}}{7} \right) \)
5. \( x \in (-\infty, \infty) \)
6. \( x \in (-\infty, -2) \cup (2, \infty) \)
7. \( x = \frac{1}{2} \)
8. \( x \in (-\infty, -\sqrt{\frac{13}{2}}) \cup (\sqrt{\frac{13}{2}}, \infty) \)
9. \( x \in (-4, -3) \cup (3, 4) \)
10. \( x \in (-\infty, -5] \cup [5, \infty) \)

\[
\boxed{
\begin{aligned}
1. & \quad (-\infty, 2] \cup [4, \infty) \\
2. & \quad (-\infty, \infty) \\
3. & \quad (-\infty, \infty) \\
4. & \quad \left( \frac{-1 - \sqrt{197}}{7}, \frac{-1 + \sqrt{197}}{7} \right) \\
5. & \quad (-\infty, \infty) \\
6. & \quad (-\infty, -2) \cup (2, \infty) \\
7. & \quad \left\{ \frac{1}{2} \right\} \\
8. & \quad (-\infty, -\sqrt{\frac{13}{2}}) \cup (\sqrt{\frac{13}{2}}, \infty) \\
9. & \quad (-4, -3) \cup (3, 4) \\
10. & \quad (-\infty, -5] \cup [5, \infty)
\end{aligned}
}
\]