imath - Grade 11 exercises re. solving quadratic inequalities - Free Printable

To solve quadratic inequalities, we generally follow these steps:

1. Rewrite the inequality in standard form (if necessary).
2. Find the roots of the corresponding equation (set the inequality to zero and solve for \( x \)).
3. Determine the intervals defined by the roots.
4. Test a point in each interval to determine where the inequality holds.
5. Consider the boundary points based on whether the inequality is strict (\( < \) or \( > \)) or non-strict (\( \leq \) or \( \geq \)).

Let's solve a few of the inequalities step by step as examples:

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Problem 1: \( x(x + 4) \leq 0 \)

#### Step 1: Find the roots
Set \( x(x + 4) = 0 \):
\[
x = 0 \quad \text{or} \quad x + 4 = 0 \implies x = -4
\]
The roots are \( x = -4 \) and \( x = 0 \).

#### Step 2: Determine intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -4), \quad (-4, 0), \quad (0, \infty)
\]

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

#### Step 4: Consider the boundary points
The inequality is \( \leq 0 \), so we include the roots \( x = -4 \) and \( x = 0 \).

#### Solution:
\[
[-4, 0]
\]

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

#### Step 1: Factorize the quadratic expression
\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]
Set \( (x - 4)(x + 1) = 0 \):
\[
x = 4 \quad \text{or} \quad x = -1
\]
The roots are \( x = -1 \) and \( x = 4 \).

#### Step 2: Determine intervals
The roots divide the real number line into three intervals:
\[
(-\infty, -1), \quad (-1, 4), \quad (4, \infty)
\]

#### Step 3: Test a point in each interval
- For \( x \in (-\infty, -1) \): Choose \( x = -2 \):
\[
(x - 4)(x + 1) = (-2 - 4)(-2 + 1) = (-6)(-1) = 6 > 0
\]
- For \( x \in (-1, 4) \): Choose \( x = 0 \):
\[
(x - 4)(x + 1) = (0 - 4)(0 + 1) = (-4)(1) = -4 < 0
\]
- For \( x \in (4, \infty) \): Choose \( x = 5 \):
\[
(x - 4)(x + 1) = (5 - 4)(5 + 1) = (1)(6) = 6 > 0
\]

#### Step 4: Consider the boundary points
The inequality is \( \leq 0 \), so we include the roots \( x = -1 \) and \( x = 4 \).

#### Solution:
\[
[-1, 4]
\]

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Problem 3: \( (x - 2)(x + 3) > 0 \)

#### Step 1: Find the roots
Set \( (x - 2)(x + 3) = 0 \):
\[
x = 2 \quad \text{or} \quad x = -3
\]
The roots are \( x = -3 \) and \( x = 2 \).

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

#### Step 3: Test a point in each interval
- For \( x \in (-\infty, -3) \): Choose \( x = -4 \):
\[
(x - 2)(x + 3) = (-4 - 2)(-4 + 3) = (-6)(-1) = 6 > 0
\]
- For \( x \in (-3, 2) \): Choose \( x = 0 \):
\[
(x - 2)(x + 3) = (0 - 2)(0 + 3) = (-2)(3) = -6 < 0
\]
- For \( x \in (2, \infty) \): Choose \( x = 3 \):
\[
(x - 2)(x + 3) = (3 - 2)(3 + 3) = (1)(6) = 6 > 0
\]

#### Step 4: Consider the boundary points
The inequality is \( > 0 \), so we exclude the roots \( x = -3 \) and \( x = 2 \).

#### Solution:
\[
(-\infty, -3) \cup (2, \infty)
\]

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Problem 4: \( (x + 1)(x - 3) > 0 \)

#### Step 1: Find the roots
Set \( (x + 1)(x - 3) = 0 \):
\[
x = -1 \quad \text{or} \quad x = 3
\]
The roots are \( x = -1 \) and \( x = 3 \).

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

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

#### Step 4: Consider the boundary points
The inequality is \( > 0 \), so we exclude the roots \( x = -1 \) and \( x = 3 \).

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

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Final Answer

\[
\boxed{[-4, 0], [-1, 4], (-\infty, -3) \cup (2, \infty), (-\infty, -1) \cup (3, \infty)}
\]