Inequalities and Modulus Functions

Inequalities look like equations but behave differently in one decisive respect: multiplying or dividing by a negative number reverses the sign. That single fact dictates the entire methodology — you never cross-multiply blindly, you bring everything to one side and analyse signs. Layered on top is the modulus function $|x|$ , which folds the number line at the origin and forces you to reason in cases (or to square away the modulus when it is safe to do so).

This lesson develops both rigorously: the definition and properties of $|x|$ , solving modulus equations and inequalities, and the sign-analysis method for rational and polynomial inequalities that is the heart of the Further-Maths treatment.

Where this sits in AQA 7367

This is the inequalities and modulus strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It extends A-Level Maths quadratic inequalities and modulus graphs, and connects to rational functions and graphs (sign of a quotient is read off the same critical points). The mechanical solving is AO1; the careful case analysis, the justification of why squaring is valid, and the rigorous handling of strict-vs-inclusive endpoints carry the AO2 reasoning marks.

Core theory: modulus and the rules of inequalities

The modulus function

The modulus (absolute value) of a real number is

$|x| = \begin{cases} x & x \ge 0, \\ -x & x < 0. \end{cases}$

Geometrically, $|x|$ is the distance of $x$ from 0 on the number line, and more generally $|x - a|$ is the distance between $x$ and $a$ . This distance reading is the fastest way to interpret simple modulus inequalities: $|x - a| < d$ means "within $d$ of $a$ ", i.e. $a - d < x < a + d$ .

Property	Statement
Non-negativity	$
Even	$
Multiplicative	$
Square link	$
Triangle inequality	$
Reverse triangle	$\big

The square link $|x|^2 = x^2$ is the licence behind the squaring method below: it lets you remove a modulus whenever both sides are known non-negative.

The two rules of inequalities

Adding/subtracting the same quantity to both sides, and multiplying/dividing by a positive number, preserve the inequality.
Multiplying/dividing by a negative number reverses it.

Because the sign of an algebraic expression like $x - 2$ is unknown, you must never multiply both sides by it. The safe strategy for any rational inequality is therefore: bring everything to one side, combine into a single fraction, and analyse the sign of that fraction from its critical points.

Why sign analysis works

The sign-table method rests on a simple fact: a product or quotient of factors changes sign only where one of the factors is zero (or, for a quotient, where the denominator is zero and the expression is undefined). Between consecutive such critical points, every factor keeps a constant sign, so the whole expression keeps a constant sign. That is why you can determine the sign on an entire interval by testing a single convenient point in it — or by tracking the sign of each factor in a table. The critical points partition the number line into finitely many intervals, and the expression is "all $+$ " or "all $-$ " on each.

A subtlety distinguishes numerator zeros from denominator zeros, and it governs the endpoints of your solution. At a numerator zero the expression equals $0$ : include it if the inequality is " $\ge$ " or " $\le$ ", exclude it if strict. At a denominator zero the expression is undefined: exclude it always, regardless of $\ge/\le$ versus $>/<$ . Getting these endpoint decisions right is where the AO2 marks concentrate, because it is precisely where careless candidates slip.

An alternative to multiplying by $x - 2$ (forbidden) that some find intuitive is to multiply by $(x-2)^2$ , which is never negative, preserving the inequality. This clears the denominator safely and turns a rational inequality into a polynomial one; you then solve the polynomial and remember to exclude $x = 2$ at the end. It is a legitimate route, though the bring-to-one-side method is usually tidier.

Worked examples (with mark scheme)

Example 1 — a modulus equation by cases

Solve $|2x - 3| = x + 1$ .

A modulus equals $\pm$ its contents, so split into two cases (and we will check the implicit requirement $x + 1 \ge 0$ at the end):

$\textbf{Case 1: } 2x - 3 = x + 1 \;\Rightarrow\; x = 4. \quad (\text{M1 set up both cases; A1})$ $\textbf{Case 2: } 2x - 3 = -(x + 1) \;\Rightarrow\; 3x = 2 \;\Rightarrow\; x = \tfrac{2}{3}. \quad (\text{A1})$

Check both in the original (essential, since squaring/casework can introduce false roots): at $x=4$ , $|5| = 5$ and $4+1 = 5$ ✓; at $x=\tfrac23$ , $\big|\tfrac43 - 3\big| = \tfrac53$ and $\tfrac23 + 1 = \tfrac53$ ✓. Both valid.

$\therefore\; x = 4 \quad\text{or}\quad x = \tfrac{2}{3}. \quad (\text{A1})$

(M1 for both cases; A1 each value; A1 for verifying and stating both solutions. Always test candidates back in the original equation.)

Example 2 — a modulus inequality by squaring

Solve $|x - 2| < |2x + 1|$ .

Both sides are non-negative, so squaring is valid and removes both moduli ( $|A|^2 = A^2$ ):

$(x - 2)^2 < (2x + 1)^2. \quad (\text{M1 square both sides})$ $x^2 - 4x + 4 < 4x^2 + 4x + 1 \;\Rightarrow\; 0 < 3x^2 + 8x - 3 = (3x - 1)(x + 3). \quad (\text{M1 rearrange and factorise})$

The quadratic $(3x-1)(x+3)$ is an upward parabola with roots $x = \tfrac13$ and $x = -3$ , so it is positive outside the roots:

$x < -3 \quad\text{or}\quad x > \tfrac{1}{3}. \quad (\text{A1})$

(M1 squaring; M1 rearranging to a factorised quadratic; A1 for the correct exterior region. Squaring is legitimate only because both sides are non-negative — state this. A useful check: at the "balance point" where $x - 2$ and $2x+1$ have equal modulus, $x = \tfrac13$ and $x=-3$ , matching the boundaries.)

Squaring is the workhorse for modulus inequalities with a modulus on each side, and it works because $|A| < |B| \iff A^2 < B^2$ — both sides being non-negative, squaring is a reversible (order-preserving) step. The justification matters for the AO2 mark, so state "both sides are non-negative" explicitly before you square. Contrast this with a modulus equation (Example 1), where squaring also works ( $|A| = |B| \iff A^2 = B^2$ ) but the by-cases method is often quicker, and with a "one-sided" inequality like $|f(x)| < g(x)$ where $g$ might be negative — there squaring is unsafe and you must argue $-g < f < g$ (which silently requires $g > 0$ ). Knowing when squaring is valid is as important as the algebra itself.

Example 3 — a rational inequality, done safely

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

Do not multiply by $x - 2$ (its sign is unknown). Bring to one side and combine:

$\frac{x + 1}{x - 2} - 1 > 0 \;\Rightarrow\; \frac{(x + 1) - (x - 2)}{x - 2} > 0 \;\Rightarrow\; \frac{3}{x - 2} > 0. \quad (\text{M1 one side; M1 single fraction})$

Since the numerator $3 > 0$ , the fraction is positive exactly when the denominator is positive:

$x - 2 > 0 \;\Rightarrow\; x > 2. \quad (\text{A1})$

(M1 for subtracting 1 rather than cross-multiplying; M1 for the single fraction $\tfrac{3}{x-2}$ ; A1 for $x > 2$ . The whole point of the topic: rearrange, never multiply by an unknown-sign expression.)

Example 4 — a rational inequality needing a sign table

Solve $\dfrac{x^2 - x - 2}{x - 1} \ge 0$ .

Factor the numerator: $x^2 - x - 2 = (x - 2)(x + 1)$ , so the expression is $\dfrac{(x-2)(x+1)}{x-1}$ . $(\text{M1 factorise})$ The critical points (zeros of numerator and denominator) are $x = -1, 1, 2$ , splitting the line into four intervals. Build a sign table:

Interval	$x+1$	$x-1$	$x-2$	quotient
$x < -1$	$-$	$-$	$-$	$-$
$-1 < x < 1$	$+$	$-$	$-$	$+$
$1 < x < 2$	$+$	$+$	$-$	$-$
$x > 2$	$+$	$+$	$+$	$+$

$(\text{M1 sign table})$ We need $\ge 0$ , so the $+$ regions, including the numerator zeros $x = -1$ and $x = 2$ (where the value is 0) but excluding $x = 1$ (where it is undefined):

$-1 \le x < 1 \quad\text{or}\quad x \ge 2. \quad (\text{A1})$

(M1 factorise; M1 a correct sign table over all critical points; A1 for the regions with correct open/closed endpoints. The endpoint logic is where AO2 marks live: numerator zeros are included for " $\ge$ ", but a denominator zero is always excluded.)

This example shows the full sign-table machine in action, and three points are worth drawing out. First, factorise completely before tabulating — the critical points are the zeros of the factors, so an unfactorised $x^2 - x - 2$ hides them. Second, the table tracks the sign of each factor, then multiplies the signs down each row; an odd number of negative factors gives a negative quotient. Third — and most examinable — the endpoints are decided factor-by-factor: $x = -1$ and $x = 2$ are numerator zeros, so the expression is $0$ there and they are included for " $\ge 0$ "; $x = 1$ is a denominator zero, so the expression is undefined and it is excluded whatever the inequality. The half-open answer $-1 \le x < 1$ captures exactly this asymmetry. A quick way to confirm a sign table is to test one point per interval (e.g. $x = 0$ gives $\tfrac{(-2)(1)}{-1} = 2 > 0$ , confirming the $+$ on $-1 < x < 1$ ).

Specimen-style exam question

(specimen-style — not from any past paper)

(a) Solve the inequality $|x - 1| \ge |x + 3|$ . (b) On a single sketch, draw $y = |x - 1|$ and $y = |x + 3|$ , and use it to confirm your answer to (a).

(a) Both sides non-negative, so square:

$(x - 1)^2 \ge (x + 3)^2 \;\Rightarrow\; x^2 - 2x + 1 \ge x^2 + 6x + 9 \;\Rightarrow\; -8x \ge 8 \;\Rightarrow\; x \le -1.$

(Dividing by $-8$ reversed the inequality — the crucial step.) So $x \le -1$ .

(b) The graphs are V-shapes with vertices at $x = 1$ and $x = -3$ ; they intersect where the distances from $1$ and $-3$ are equal, i.e. at the midpoint $x = -1$ . To the left of $x = -1$ , the point is closer to $-3$ than to $1$ , so $|x-1| \ge |x+3|$ there — confirming $x \le -1$ .

graph LR
  A["x = −1 is the midpoint of −3 and 1"] --> B["left of −1: closer to −3 ⇒ |x−1| ≥ |x+3|"]
  A --> C["right of −1: closer to 1 ⇒ |x−1| < |x+3|"]

Inequalities and Modulus Functions

Inequalities and Modulus Functions

Where this sits in AQA 7367

Core theory: modulus and the rules of inequalities

The modulus function

The two rules of inequalities

Why sign analysis works

Worked examples (with mark scheme)

Example 1 — a modulus equation by cases

Example 2 — a modulus inequality by squaring

Example 3 — a rational inequality, done safely

Example 4 — a rational inequality needing a sign table

Specimen-style exam question

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