What is wrong interval for a quadratic?

(x - a)(x - b) 0 holds outside. Sketch to avoid reversing this.

What is sign error across a repeated root?

At a double root the sign does not change; only odd-multiplicity roots flip the sign.

Solve x^2 - 5x + 6 ≥ 0. [3 marks]

Explain why you should not multiply both sides of 1x < 2 by x. [2 marks]

SingaporeMathsSyllabus dot point

How do we solve polynomial, rational and modulus inequalities reliably?

Solve quadratic, polynomial and rational inequalities algebraically and graphically, using a sign analysis and respecting the sign of any denominator

A focused answer to the H2 Mathematics outcome on inequalities. Solving quadratic and higher polynomial inequalities by sign analysis, handling rational inequalities without cross-multiplying carelessly, and reading solutions off a graph.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to solve quadratic, higher-degree polynomial and rational inequalities, both algebraically by sign analysis and graphically, while being careful never to multiply across by an expression whose sign is unknown.

The answer

Quadratic inequalities

Bring everything to one side so it reads $\text{expression} \lessgtr 0$ , factorise, find the roots (the critical values), and decide the sign between and outside them. For an upward parabola the expression is negative between the roots and positive outside; reverse for a downward parabola. A quick sketch settles it.

Polynomial inequalities by sign analysis

For higher-degree expressions, find all the roots, mark them on a number line, and determine the sign of the product in each interval (the sign flips at a simple root and stays the same across a repeated root of even multiplicity). Read off the intervals satisfying the inequality.

Rational inequalities

The critical trap: never cross-multiply by a denominator whose sign you do not know, because multiplying an inequality by a negative quantity reverses it. Two safe approaches:

Move everything to one side, combine into a single fraction, and do a sign analysis of numerator and denominator together. The denominator's zeros are critical values but are always excluded.
Multiply by the square of the denominator (which is positive), reducing to a polynomial inequality, then exclude the points where the denominator is zero.

Reading inequalities from a graph

The solution of $\mathrm{f}(x) > \mathrm{g}(x)$ is the set of $x$ where the graph of $\mathrm{f}$ lies above the graph of $\mathrm{g}$ . Finding the intersection points and reading the regions is often the fastest route and is exactly how the graphing calculator helps.

Worked example

Solve the inequality $\dfrac{x + 4}{x - 1} \leq 2$ .

Step 1: Move everything to one side

\frac{x + 4}{x - 1} - 2 \leq 0

Step 2: Combine into a single fraction

\frac{x + 4 - 2(x - 1)}{x - 1} = \frac{x + 4 - 2x + 2}{x - 1} = \frac{6 - x}{x - 1} \leq 0

Step 3: Identify critical values

Numerator zero at $x = 6$ ; denominator zero at $x = 1$ (excluded).

Step 4: Sign analysis

For $x < 1$ (say $x = 0$ ): $\dfrac{6}{-1} < 0$ , included. For $1 < x < 6$ (say $x = 2$ ): $\dfrac{4}{1} > 0$ , excluded. For $x > 6$ (say $x = 7$ ): $\dfrac{-1}{6} < 0$ , included.

Step 5: Endpoints and solution

$x = 6$ gives $0$ , satisfying $\leq 0$ , so included; $x = 1$ excluded. Solution: $x < 1$ or $x \geq 6$ .

Examples in context

Example 1. A profit threshold. A profit model $P(x) = -x^2 + 10x - 16$ is positive when $-x^2 + 10x - 16 > 0$ , that is $(x - 2)(x - 8) < 0$ , so $2 < x < 8$ . The business is profitable only for production levels strictly between $2$ and $8$ units.

Example 2. Comparing two rates. To find where $\dfrac{1}{x} > \dfrac{1}{x - 3}$ , move to one side: $\dfrac{1}{x} - \dfrac{1}{x-3} = \dfrac{-3}{x(x-3)} > 0$ , so $x(x - 3) < 0$ , giving $0 < x < 3$ . Cross-multiplying would have produced a wrong answer.

Try this

Q1. Solve $x^2 - 5x + 6 \geq 0$ . [3 marks]

Cue. $(x - 2)(x - 3) \geq 0$ , so $x \leq 2$ or $x \geq 3$ .

Q2. Solve $\dfrac{2}{x - 1} > 1$ . [3 marks]

Cue. $\dfrac{2 - (x - 1)}{x - 1} = \dfrac{3 - x}{x - 1} > 0$ , so $1 < x < 3$ .

Q3. Explain why you should not multiply both sides of $\dfrac{1}{x} < 2$ by $x$ . [2 marks]

Cue. $x$ may be negative, which would reverse the inequality; the sign is unknown, so rearrange to one side instead.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksSolve the inequality

\dfrac{x - 1}{x + 2} \geq 0

Show worked answer →

The expression is zero or undefined at the critical values $x = 1$ (numerator zero) and $x = -2$ (denominator zero, excluded).

Test the three regions:
For $x < -2$ (say $x = -3$ ): $\dfrac{-4}{-1} = 4 > 0$ , included.
For $-2 < x < 1$ (say $x = 0$ ): $\dfrac{-1}{2} < 0$ , excluded.
For $x > 1$ (say $x = 2$ ): $\dfrac{1}{4} > 0$ , included.

The numerator zero $x = 1$ is included (gives $0$ ); $x = -2$ is excluded (undefined).

Solution: $x < -2$ or $x \geq 1$ .

Markers reward identifying both critical values, a correct sign analysis, and including or excluding the endpoints correctly.

Original4 marksSolve the inequality

x^2 < 2x + 3

algebraically.

Show worked answer →

Bring all terms to one side: $x^2 - 2x - 3 < 0$ .

Factorise: $(x - 3)(x + 1) < 0$ .

The product is negative between the roots, so $-1 < x < 3$ .

A sketch of the upward parabola confirms it is below the $x$ -axis exactly between its roots $x = -1$ and $x = 3$ .

Markers reward rearranging to one side, factorising, and the correct interval with strict inequalities.