Quadratic Inequalities Revision Notes for Grade 11 NSC Matric Mathematics

Quadratic Inequalities

What are quadratic inequalities?

A quadratic inequality is an inequality that involves a quadratic expression. Unlike quadratic equations where we find exact values, quadratic inequalities ask us to find ranges of values that satisfy the given condition.

Quadratic inequalities can take four main forms:

$ax^2 + bx + c > 0$ (greater than zero)
$ax^2 + bx + c \geq 0$ (greater than or equal to zero)
$ax^2 + bx + c < 0$ (less than zero)
$ax^2 + bx + c \leq 0$ (less than or equal to zero)

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The key difference between quadratic equations and inequalities is that equations give us specific x-values as solutions, while inequalities give us ranges or intervals of x-values that make the inequality true.

How to solve quadratic inequalities

To solve a quadratic inequality, we need to determine which parts of the graph of the quadratic function lie above or below the x-axis. The most reliable approach uses a systematic method with a table of signs.

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Why use a table of signs?

A table of signs helps us organize our work systematically and ensures we don't miss any intervals or make sign errors. This method works consistently for all types of quadratic inequalities.

Step-by-step method:

Step 1: Factorise the quadratic expression

Write the quadratic in factorised form if possible
If factorisation is difficult, use the quadratic formula to find the roots first

Step 2: Find the critical values

Critical values are the x-values where the expression equals zero
These are the roots of the quadratic equation
Set each factor equal to zero to find these values

Step 3: Create a table of signs

Use the critical values to divide the number line into intervals
Test the sign of each factor in each interval
Multiply the signs to find where the whole expression is positive or negative

Step 4: Sketch a rough graph (optional)

This helps visualise the solution
Remember: if $a > 0$ , the parabola opens upward (U-shape)
If $a < 0$ , the parabola opens downward (∩-shape)

Step 5: Write the final answer

Express your solution in interval notation or inequality form
Represent the solution on a number line

Worked examples

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Worked Example 1: Standard quadratic inequality

Question: Solve $x^2 - 5x + 6 \geq 0$

Solution:

Step 1: Factorise the quadratic $x^2 - 5x + 6 = (x - 3)(x - 2)$

So we need to solve: $(x - 3)(x - 2) \geq 0$

Step 2: Find critical values Setting each factor to zero: $x = 2$ and $x = 3$

Step 3: Complete the table of signs

Table

From this table, we can see that $f(x) \geq 0$ when $x \leq 2$ or $x \geq 3$ .

Step 4: Sketch the graph

The graph shows that the parabola is above or on the x-axis for $x \leq 2$ or $x \geq 3$ .

Step 5: Final answer

$x^2 - 5x + 6 \geq 0$ for $x \leq 2$ or $x \geq 3$

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Worked Example 2: Perfect square case

Question: Solve $4x^2 - 4x + 1 \leq 0$

Solution:

Step 1: Factorise the quadratic $4x^2 - 4x + 1 = (2x - 1)^2$

So we need: $(2x - 1)^2 \leq 0$

Step 2: Find critical values
Setting $(2x - 1)^2 = 0$ gives us $2x - 1 = 0$ , so $x = \frac{1}{2}$

Since $(2x - 1)^2$ is always positive or zero (being a perfect square), it can only equal zero at $x = \frac{1}{2}$ .

Step 3: Analysis Since a square of any real number is never negative, $(2x - 1)^2 < 0$ has no solution. $(2x - 1)^2 = 0$ only when $x = \frac{1}{2}$ .

Final answer: $x = \frac{1}{2}$

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Worked Example 3: Using the quadratic formula

Question: Solve $x^2 + 3x - 5 < 0$

Solution:

Step 1: Check if factorisation is possible

The expression $x^2 + 3x - 5$ cannot be easily factorised, so we use the quadratic formula.

Step 2: Find the roots using the quadratic formula

For $x^2 + 3x - 5 = 0$ :

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-5)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 20}}{2} = \frac{-3 \pm \sqrt{29}}{2}$

$x_1 = \frac{-3 - \sqrt{29}}{2} \approx -4.2$ and $x_2 = \frac{-3 + \sqrt{29}}{2} \approx 1.2$

Step 3: Create the table of signs

Table

Step 4: Analyse the graph

Since $a = 1 > 0$ , the parabola opens upward. The quadratic is negative between the two roots.

Step 5: Final answer

$x^2 + 3x - 5 < 0$ for $-4.2 < x < 1.2$ (approximately)

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Worked Example 4: Quadratic inequalities with fractions

Question: Solve $\frac{2}{x + 3} \leq \frac{1}{x - 3}$ where $x \neq \pm 3$

Solution:

Step 1: Rearrange the inequality

$\frac{2}{x + 3} - \frac{1}{x - 3} \leq 0$

Step 2: Find a common denominator

$\frac{2(x - 3) - 1(x + 3)}{(x + 3)(x - 3)} \leq 0$

$\frac{2x - 6 - x - 3}{(x + 3)(x - 3)} \leq 0$

$\frac{x - 9}{(x + 3)(x - 3)} \leq 0$

Step 3: Find critical values

The critical values are: $x = -3$ , $x = 3$ , and $x = 9$

Note: $x = -3$ and $x = 3$ make the denominator zero, so they're restrictions, not part of the solution.

Step 4: Complete the table of signs

Table

Step 5: Final answer

The solution is: $x < -3$ or $3 < x \leq 9$

Important exam tips

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Common Pitfalls to Avoid:

Sign changes: When multiplying or dividing an inequality by a negative number, remember to flip the inequality sign.
Perfect squares: Remember that $(expression)^2 \geq 0$ always, and $(expression)^2 = 0$ only when the expression equals zero.
Restrictions: When dealing with fractions, always identify values that make denominators zero - these cannot be part of your solution.
Boundary values: Pay attention to whether boundary points are included (≤, ≥) or excluded (<, >) in your final answer.

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Key Points to Remember:

Quadratic inequalities involve finding ranges of x-values rather than specific solutions
Use a table of signs to systematically determine where the expression is positive or negative
Critical values are where the expression equals zero - these divide your number line into intervals
Always check restrictions when fractions are involved in the inequality
Perfect squares are never negative - they can only be positive or zero

Quadratic Inequalities (Grade 11 NSC Matric Mathematics): Revision Notes