Inequalities

The CSEC syllabus focuses on two types of inequality beyond the linear work covered in General Mathematics: quadratic inequalities and rational inequalities with linear factors. Both are solved by the same underlying strategy: find critical values, then determine the sign of the expression in each region.

The Key Idea: Sign Analysis

An inequality like $f(x) > 0$ asks where the expression $f(x)$ is positive. The sign of $f(x)$ can only change at points where $f(x) = 0$ or where $f(x)$ is undefined. These are the critical values. Between consecutive critical values, the sign is constant, so you only need to test one point per region.

Quadratic Inequalities

Method:

Rearrange so the right side is zero.
Solve $ax^2 + bx + c = 0$ to find the critical values (the roots of the equality).
State the critical values, then sketch the parabola with those roots marked on the x-axis. Shade the region the inequality describes.
Read off the solution and write it in set builder notation.

For $a > 0$ (upward parabola) with critical values $\alpha < \beta$ : the expression is negative between the roots and positive outside them. For $a < 0$ the regions reverse.

Example

Solve $x^2 - x - 6 < 0$ .

Solve: $(x - 3)(x + 2) = 0$

Critical values: $x = -2$ , $x = 3$

Sketch: parabola opens upward ( $a = 1 > 0$ ). It lies below the x-axis between the roots, so shade that region.

$\lbrace x \in \mathbb{R} : -2 < x < 3 \rbrace$

Example

Solve $x^2 - 9 \geq 0$ .

Solve: $(x - 3)(x + 3) = 0$

Critical values: $x = -3$ , $x = 3$

Sketch: parabola opens upward. It lies above (or on) the x-axis outside the roots, so shade those two outer regions. Endpoints are included because the inequality is $\geq$ .

$\lbrace x \in \mathbb{R} : x \leq -3 \rbrace \cup \lbrace x \in \mathbb{R} : x \geq 3 \rbrace$

Using a Sign Diagram

A sign diagram tracks the sign of each factor across every region. It is especially clear when the expression is already in factorised form.

Example

Solve $(x + 1)(x - 4) \leq 0$ .

Critical values: $x = -1$ , $x = 4$

Region	$x + 1$	$x - 4$	Product
$x < -1$	$-$	$-$	$+$
$-1 < x < 4$	$+$	$-$	$-$
$x > 4$	$+$	$+$	$+$

The product is $\leq 0$ in the middle region. Both endpoints satisfy the equality, so they are included.

$\lbrace x \in \mathbb{R} : -1 \leq x \leq 4 \rbrace$

Exam Tip

Never multiply both sides by just the denominator to clear a fraction — its sign depends on $x$ , so the inequality direction may flip. Multiply by the denominator squared instead: a square is always positive, so the direction never changes.

Rational Inequalities with Linear Factors

The syllabus specifies inequalities of the form $\dfrac{ax + b}{cx + d} > 0$ (or $\geq 0$ , $< 0$ , $\leq 0$ ).

Method: multiply both sides by $(cx + d)^2$ . Because a square is always positive, the inequality direction does not change. This turns the rational inequality into a quadratic:

$\frac{ax + b}{cx + d} \gtreqqless 0 \implies (ax + b)(cx + d) \gtreqqless 0$

Solve the resulting quadratic inequality using the parabola method. At the end, exclude any $x$ that makes the original denominator zero — the fraction is undefined there even if the quadratic would include it.

Example

Solve $\dfrac{x - 1}{x + 2} > 0$ .

Multiply both sides by $(x + 2)^2$ :

$(x - 1)(x + 2) > 0$

Roots: $x = 1$ and $x = -2$ . The parabola opens upward ( $a > 0$ ), so the expression is positive outside the roots.

Solution from the quadratic: $x < -2$ or $x > 1$ .

Exclude $x = -2$ (denominator zero, fraction undefined). Both regions already exclude it, so no adjustment is needed.

$\lbrace x \in \mathbb{R} : x < -2 \rbrace \cup \lbrace x \in \mathbb{R} : x > 1 \rbrace$

Example

Solve $\dfrac{2x + 3}{x - 5} \leq 0$ .

Multiply both sides by $(x - 5)^2$ :

$(2x + 3)(x - 5) \leq 0$

Roots: $x = -\dfrac{3}{2}$ and $x = 5$ . The parabola opens upward, so the expression is $\leq 0$ between the roots.

Solution from the quadratic: $-\dfrac{3}{2} \leq x \leq 5$ .

Exclude $x = 5$ (denominator zero). Change the right endpoint to open.

$\left\lbrace x \in \mathbb{R} : -\frac{3}{2} \leq x < 5 \right\rbrace$

Notation Reference

Set builder notation (standard form for exam solutions):

Situation	Set builder form
Connected range, strict	$\lbrace x \in \mathbb{R} : a < x < b \rbrace$
Connected range, closed	$\lbrace x \in \mathbb{R} : a \leq x \leq b \rbrace$
Mixed endpoints	$\lbrace x \in \mathbb{R} : a \leq x < b \rbrace$
Two separate ranges	$\lbrace x \in \mathbb{R} : x < a \rbrace \cup \lbrace x \in \mathbb{R} : x > b \rbrace$

The symbol $\in$ means "is a member of" and $\mathbb{R}$ denotes the set of real numbers.

Interval notation (alternative, also acceptable):

Notation	Meaning
$(a, b)$	$a < x < b$
$[a, b]$	$a \leq x \leq b$
$[a, b)$	$a \leq x < b$
$(-\infty, a)$	$x < a$
$A \cup B$	$x$ satisfies $A$ or $B$

Remember

Endpoint inclusion: a critical value is included (closed boundary, $\leq$ or $\geq$ ) when the inequality is not strict and the expression is defined there. It is excluded (open boundary) when the inequality is strict, or when it is a denominator-zero point (fraction undefined).

The Key Idea: Sign Analysis

Quadratic Inequalities

Method:

Rearrange so the right side is zero.
Solve $ax^2 + bx + c = 0$ to find the critical values (the roots of the equality).
State the critical values, then sketch the parabola with those roots marked on the x-axis. Shade the region the inequality describes.
Read off the solution and write it in set builder notation.

For $a > 0$ (upward parabola) with critical values $\alpha < \beta$ : the expression is negative between the roots and positive outside them. For $a < 0$ the regions reverse.

Example

Solve $x^2 - x - 6 < 0$ .

Solve: $(x - 3)(x + 2) = 0$

Critical values: $x = -2$ , $x = 3$

Sketch: parabola opens upward ( $a = 1 > 0$ ). It lies below the x-axis between the roots, so shade that region.

$\lbrace x \in \mathbb{R} : -2 < x < 3 \rbrace$

Example

Solve $x^2 - 9 \geq 0$ .

Solve: $(x - 3)(x + 3) = 0$

Critical values: $x = -3$ , $x = 3$

Sketch: parabola opens upward. It lies above (or on) the x-axis outside the roots, so shade those two outer regions. Endpoints are included because the inequality is $\geq$ .

$\lbrace x \in \mathbb{R} : x \leq -3 \rbrace \cup \lbrace x \in \mathbb{R} : x \geq 3 \rbrace$

Using a Sign Diagram

A sign diagram tracks the sign of each factor across every region. It is especially clear when the expression is already in factorised form.

Example

Solve $(x + 1)(x - 4) \leq 0$ .

Critical values: $x = -1$ , $x = 4$

Region	$x + 1$	$x - 4$	Product
$x < -1$	$-$	$-$	$+$
$-1 < x < 4$	$+$	$-$	$-$
$x > 4$	$+$	$+$	$+$

The product is $\leq 0$ in the middle region. Both endpoints satisfy the equality, so they are included.

$\lbrace x \in \mathbb{R} : -1 \leq x \leq 4 \rbrace$

Exam Tip

Rational Inequalities with Linear Factors

The syllabus specifies inequalities of the form $\dfrac{ax + b}{cx + d} > 0$ (or $\geq 0$ , $< 0$ , $\leq 0$ ).

Method: multiply both sides by $(cx + d)^2$ . Because a square is always positive, the inequality direction does not change. This turns the rational inequality into a quadratic:

$\frac{ax + b}{cx + d} \gtreqqless 0 \implies (ax + b)(cx + d) \gtreqqless 0$

Example

Solve $\dfrac{x - 1}{x + 2} > 0$ .

Multiply both sides by $(x + 2)^2$ :

$(x - 1)(x + 2) > 0$

Roots: $x = 1$ and $x = -2$ . The parabola opens upward ( $a > 0$ ), so the expression is positive outside the roots.

Solution from the quadratic: $x < -2$ or $x > 1$ .

Exclude $x = -2$ (denominator zero, fraction undefined). Both regions already exclude it, so no adjustment is needed.

$\lbrace x \in \mathbb{R} : x < -2 \rbrace \cup \lbrace x \in \mathbb{R} : x > 1 \rbrace$

Example

Solve $\dfrac{2x + 3}{x - 5} \leq 0$ .

Multiply both sides by $(x - 5)^2$ :

$(2x + 3)(x - 5) \leq 0$

Roots: $x = -\dfrac{3}{2}$ and $x = 5$ . The parabola opens upward, so the expression is $\leq 0$ between the roots.

Solution from the quadratic: $-\dfrac{3}{2} \leq x \leq 5$ .

Exclude $x = 5$ (denominator zero). Change the right endpoint to open.

$\left\lbrace x \in \mathbb{R} : -\frac{3}{2} \leq x < 5 \right\rbrace$

Notation Reference

Set builder notation (standard form for exam solutions):

Situation	Set builder form
Connected range, strict	$\lbrace x \in \mathbb{R} : a < x < b \rbrace$
Connected range, closed	$\lbrace x \in \mathbb{R} : a \leq x \leq b \rbrace$
Mixed endpoints	$\lbrace x \in \mathbb{R} : a \leq x < b \rbrace$
Two separate ranges	$\lbrace x \in \mathbb{R} : x < a \rbrace \cup \lbrace x \in \mathbb{R} : x > b \rbrace$

The symbol $\in$ means "is a member of" and $\mathbb{R}$ denotes the set of real numbers.

Interval notation (alternative, also acceptable):

Notation	Meaning
$(a, b)$	$a < x < b$
$[a, b]$	$a \leq x \leq b$
$[a, b)$	$a \leq x < b$
$(-\infty, a)$	$x < a$
$A \cup B$	$x$ satisfies $A$ or $B$

Remember