Algebraic Inequalities

This lesson covers algebraic inequalities as required by the AQA A-Level Mathematics specification (7357). You must be able to solve linear, quadratic, and rational inequalities, express solutions using set notation and interval notation, and represent solutions on a number line. Inequalities appear throughout pure mathematics and are essential for finding domains, ranges, and conditions for convergence.

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Linear Inequalities

A linear inequality involves a linear expression and is solved in the same way as a linear equation, with one crucial difference:

When you multiply or divide both sides by a negative number, you must reverse the inequality sign.

Example: Solve 3x − 7 > 5.

3x > 12
x > 4

Solution: x > 4 or in set notation {x : x > 4} or interval notation (4, ∞).

Example: Solve 2 − 5x ≤ 17.

−5x ≤ 15
x ≥ −3       (inequality reversed because we divided by −5)

Double Inequalities

Example: Solve −3 < 2x + 1 ≤ 9.

−3 < 2x + 1 ≤ 9
−4 < 2x ≤ 8
−2 < x ≤ 4

Solution: {x : −2 < x ≤ 4} or (−2, 4].

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Quadratic Inequalities

To solve a quadratic inequality:

1. Rearrange so one side is zero.
2. Factorise (or find the roots using the formula).
3. Sketch the parabola to determine where the expression is positive or negative.
4. Read off the solution from the sketch.

Method

For a quadratic ax² + bx + c with roots α and β (where α < β):

Inequality	Solution (if a > 0)
ax² + bx + c > 0	x < α or x > β
ax² + bx + c ≥ 0	x ≤ α or x ≥ β
ax² + bx + c < 0	α < x < β
ax² + bx + c ≤ 0	α ≤ x ≤ β

Example: Solve x² − 5x + 6 > 0.

Factorise: (x − 2)(x − 3) > 0. Roots are x = 2 and x = 3.

The parabola is U-shaped (coefficient of x² is positive). It is above the x-axis (positive) when x < 2 or x > 3.

Solution: x < 2 or x > 3, i.e. {x : x < 2} ∪ {x : x > 3} or (−∞, 2) ∪ (3, ∞).

Example: Solve x² + 2x − 8 ≤ 0.

Factorise: (x + 4)(x − 2) ≤ 0. Roots are x = −4 and x = 2.

The parabola is U-shaped and is below or on the x-axis between the roots.

Solution: −4 ≤ x ≤ 2, i.e. [−4, 2].

Example: Solve 2x² + x − 3 < 0.

Factorise: (2x + 3)(x − 1) < 0. Roots are x = −3/2 and x = 1.

U-shaped parabola is negative between the roots.

Solution: −3/2 < x < 1.

Exam Tip: Always sketch the parabola when solving quadratic inequalities. This makes it clear which regions satisfy the inequality. Remember: a positive leading coefficient gives a U-shape; a negative leading coefficient gives an ∩-shape.

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Quadratic Inequalities with No Real Roots

If the discriminant b² − 4ac < 0 and the leading coefficient is positive, then ax² + bx + c > 0 for all real x (the parabola never crosses the x-axis).

Example: Show that x² + x + 1 > 0 for all real x.

Discriminant = 1² − 4(1)(1) = 1 − 4 = −3 < 0

Since the discriminant is negative and the coefficient of x² is positive, the parabola is always above the x-axis. Therefore x² + x + 1 > 0 for all x ∈ ℝ.

Alternatively, completing the square: x² + x + 1 = (x + 1/2)² + 3/4 ≥ 3/4 > 0 for all x.

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Rational Inequalities (Inequalities Involving Fractions)

For inequalities involving algebraic fractions, never multiply both sides by an expression containing x (because you don't know whether it's positive or negative). Instead:

1. Bring everything to one side so the other side is zero.
2. Combine into a single fraction.
3. Find the critical values (where the numerator or denominator is zero).
4. Use a sign diagram or test values in each interval.

Example: Solve (x + 1)/(x − 2) > 0.

Critical values: x = −1 (numerator = 0) and x = 2 (denominator = 0).

Sign analysis:

Interval	(x + 1)	(x − 2)	Fraction
x < −1	−	−	+
−1 < x < 2	+	−	−
x > 2	+	+	+

The fraction is positive when x < −1 or x > 2.

Solution: x < −1 or x > 2, i.e. (−∞, −1) ∪ (2, ∞).

Note: x ≠ 2 because the denominator cannot be zero.

Example: Solve (2x − 1)/(x + 3) ≤ 1.

(2x − 1)/(x + 3) − 1 ≤ 0
(2x − 1 − (x + 3))/(x + 3) ≤ 0
(x − 4)/(x + 3) ≤ 0

Critical values: x = 4 and x = −3.

Interval	(x − 4)	(x + 3)	Fraction
x < −3	−	−	+
−3 < x < 4	−	+	−
x > 4	+	+	+

The fraction is ≤ 0 in the interval −3 < x ≤ 4 (note x = 4 is included because ≤, but x = −3 is excluded because it makes the denominator zero).

Solution: {x : −3 < x ≤ 4}.

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Set Notation and Interval Notation

Solutions to inequalities should be expressed using proper notation:

Notation	Meaning
{x : x > 3}	Set builder notation: the set of all x such that x > 3
(3, ∞)	Interval notation: open bracket means 3 is not included
[−2, 5]	Closed brackets: both endpoints included
(−∞, 1) ∪ (4, ∞)	Union of two intervals

Number Line Representation

On a number line:

• An open circle (○) means the endpoint is not included (strict inequality < or >).
• A filled circle (●) means the endpoint is included (≤ or ≥).
• Shade the region that satisfies the inequality.

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Inequalities and the Discriminant

The discriminant is often combined with inequalities in exam questions.

Example: Find the values of k for which x² + kx + 9 = 0 has two distinct real roots.

For two distinct real roots, the discriminant must be positive:

b² − 4ac > 0
k² − 36 > 0
(k − 6)(k + 6) > 0

Solution: k < −6 or k > 6.

Example: Find the values of p for which 2x² + px + 8 = 0 has no real roots.

p² − 64 < 0
(p − 8)(p + 8) < 0
−8 < p < 8

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Summary

• Solve linear inequalities like equations, but reverse the sign when multiplying or dividing by a negative.
• For quadratic inequalities, find roots, sketch the parabola, and read off the solution.
• For rational inequalities, bring to one side, combine fractions, and use a sign diagram.
• Never multiply both sides of an inequality by an expression that could be negative.
• Express solutions in set notation or interval notation, and represent on a number line.
• The discriminant combined with inequalities is a common exam technique.

Exam Tip: The most common error with quadratic inequalities is writing the wrong type of solution — e.g., writing α < x < β when it should be x < α or x > β. Always sketch the parabola. For rational inequalities, never cross-multiply blindly; always rearrange to get zero on one side first and use a sign diagram.

A-Level Deep Dive: Algebraic Inequalities

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section B (Algebra and Functions) covers linear and quadratic inequalities in a single variable and interpret such inequalities graphically, including inequalities with brackets and fractions; express solutions through correct use of 'and' and 'or', or through set notation (refer to the official specification document for exact wording). Although embedded in Section B, inequality reasoning recurs across the AQA papers: in Section H (Differentiation) when classifying stationary points using $f''(x) > 0$f′′(x)>0 or $f''(x) < 0$f′′(x)<0; in Section I (Integration) when areas are computed only over intervals where the integrand is non-negative; and in Section L (Numerical methods) when bracketing root-containing intervals using the change-of-sign principle. The AQA formula booklet does not list any inequality manipulation rules — they must be remembered and used fluently.

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Worked example with full mark scheme

Question (8 marks):

Solve the inequality $\dfrac{x - 1}{x + 2} > 3$x+2x−1​>3, giving your answer in interval notation. (8)

Solution with mark scheme:

Step 1 — recognise that you cannot multiply by $(x + 2)$(x+2) directly.

The denominator $(x + 2)$(x+2) may be positive or negative, and multiplying an inequality by a negative quantity reverses the inequality sign. The safe technique is to bring everything to one side and form a single rational expression.

M1 — recognising that direct cross-multiplication is invalid; rearranging to $\dfrac{x - 1}{x + 2} - 3 > 0$x+2x−1​−3>0.

Step 2 — combine into a single fraction over a common denominator.

$\dfrac{x - 1}{x + 2} - 3 = \dfrac{x - 1 - 3(x + 2)}{x + 2} = \dfrac{x - 1 - 3x - 6}{x + 2} = \dfrac{-2x - 7}{x + 2}$x+2x−1​−3=x+2x−1−3(x+2)​=x+2x−1−3x−6​=x+2−2x−7​

M1 — correct combination over the common denominator.

A1 — correct simplified numerator $-2x - 7$−2x−7.

Step 3 — restate the inequality.

$\dfrac{-2x - 7}{x + 2} > 0$x+2−2x−7​>0

Equivalently, multiplying numerator and denominator by $-1$−1 (and reversing the inequality once because the fraction's value is negated):

$\dfrac{2x + 7}{x + 2} < 0$x+22x+7​<0

M1 — manipulating to a clean form whose sign analysis is straightforward.

Step 4 — sign analysis via critical values.

The fraction $\dfrac{2x + 7}{x + 2}$x+22x+7​ changes sign only where the numerator is zero ($x = -\tfrac{7}{2}$x=−27​) or the denominator is zero ($x = -2$x=−2, where the expression is undefined).

Test signs on the three intervals:

• $x < -\tfrac{7}{2}$x<−27​ (e.g. $x = -4$x=−4): numerator $2(-4) + 7 = -1 < 0$2(−4)+7=−1<0; denominator $-4 + 2 = -2 < 0$−4+2=−2<0; fraction positive.
• $-\tfrac{7}{2} < x < -2$−27​<x<−2 (e.g. $x = -3$x=−3): numerator $2(-3) + 7 = 1 > 0$2(−3)+7=1>0; denominator $-3 + 2 = -1 < 0$−3+2=−1<0; fraction negative.
• $x > -2$x>−2 (e.g. $x = 0$x=0): numerator $7 > 0$7>0; denominator $2 > 0$2>0; fraction positive.

M1 — identifying the two critical values and partitioning the real line.

A1 — correct sign on each subinterval.

Step 5 — write the solution set.

The fraction is strictly less than zero on $-\tfrac{7}{2} < x < -2$−27​<x<−2. The endpoint $x = -\tfrac{7}{2}$x=−27​ is excluded because the inequality is strict (the fraction equals zero there, not less than zero); the endpoint $x = -2$x=−2 is excluded because the expression is undefined.

In interval notation: $x \in \left(-\tfrac{7}{2}, -2\right)$x∈(−27​,−2).

A1 — correct interval, both endpoints open.

A1 — final answer presented in the requested form (interval notation), with explicit justification of why each endpoint is excluded.

Total: 8 marks (M4 A4, split as shown). The defining features of an A* answer here are (i) refusing to cross-multiply by an unsigned denominator and (ii) explicitly distinguishing "open because strict" from "open because undefined" — examiners reward this precision.

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The function $f$f is defined by $f(x) = 2x^2 + 5x - 3$f(x)=2x2+5x−3.

(a) Find the values of $x$x for which $f(x) \leq 0$f(x)≤0, giving your answer in set-builder notation. (4)

(b) Hence state the set of values of $x$x for which $f(x) > 0$f(x)>0. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — factorising $2x^2 + 5x - 3 = (2x - 1)(x + 3)$2x2+5x−3=(2x−1)(x+3).
• A1 (AO1.1b) — identifying critical values $x = \tfrac{1}{2}$x=21​ and $x = -3$x=−3.
• M1 (AO2.1) — recognising that the parabola opens upward (positive leading coefficient $2$2), so $f(x) \leq 0$f(x)≤0 between the roots.
• A1 (AO2.5) — solution $\{x : -3 \leq x \leq \tfrac{1}{2}\}${x:−3≤x≤21​}, with closed endpoints (the inequality is non-strict).

(b)

• M1 (AO2.2a) — recognising that $f(x) > 0$f(x)>0 is the complement of $f(x) \leq 0$f(x)≤0 (excluding the endpoints, since $f = 0$f=0 there).
• A1 (AO2.5) — solution $\{x : x < -3\} \cup \{x : x > \tfrac{1}{2}\}${x:x<−3}∪{x:x>21​}.

Total: 6 marks split AO1 = 2, AO2 = 4. This is an AO2-heavy question — AQA uses quadratic inequalities to test interpretation and notation precision rather than raw factorisation.

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Synoptic links

Connects to:

1. Section B — Linear and quadratic inequalities (the simpler cousins): every rational inequality reduces, after a careful "bring to one side" step, to a sign-analysis question on a polynomial expression. Mastering linear ($ax + b > 0$ax+b>0) and quadratic ($ax^2 + bx + c > 0$ax2+bx+c>0) inequalities is therefore a prerequisite for all higher-level inequality work.

2. Section B — Modulus / absolute-value inequalities: $|x - 3| < 5$∣x−3∣<5 is equivalent to $-5 < x - 3 < 5$−5<x−3<5, i.e. $-2 < x < 8$−2<x<8. Modulus inequalities are routinely solved by squaring (when both sides are non-negative) or by splitting into cases — the same case-analysis discipline as rational inequalities.

3. Section E — Binomial expansion validity range: the expansion $(1 + x)^n$(1+x)n for non-integer $n$n converges only when $|x| < 1$∣x∣<1. Establishing this validity range is an inequality. Composite expansions like $(1 + 2x)^{-1/2}$(1+2x)−1/2 require $|2x| < 1$∣2x∣<1, i.e. $|x| < \tfrac{1}{2}$∣x∣<21​.

4. Section H — Stationary points and curve sketching: classifying a stationary point requires the inequality $f''(x) > 0$f′′(x)>0 (minimum) or $f''(x) < 0$f′′(x)<0 (maximum). Sketching curves in inequality regions ($y > f(x)$y>f(x) shading) re-uses exactly the sign-analysis logic developed here.

5. Section I — Integration of signed functions: the area between a curve and the $x$x-axis is $\int_a^b |f(x)|\,dx$∫ab​∣f(x)∣dx, but the integral $\int_a^b f(x)\,dx$∫ab​f(x)dx requires the integrand to be sign-tracked across subintervals where $f(x) > 0$f(x)>0 or $f(x) < 0$f(x)<0. Setting up these subintervals is an inequality-solving exercise.

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Mark-scheme literacy

Algebraic inequality questions on 7357 Paper 1 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Factorising, identifying critical values, applying inequality manipulation rules correctly
AO2 (reasoning / interpretation)	30–40%	Choosing a valid manipulation (not cross-multiplying by unsigned denominator), interpreting the sign of the leading coefficient, presenting solutions in the requested notation (interval / set-builder / number line)
AO3 (problem-solving)	0–15%	Modelling contexts where an inequality arises naturally (e.g. "for what values of $k$k does the equation have two distinct real roots?")