Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Linear and quadratic inequalities

Pure content (sections A–J) · Subtopic 4 of 66

Linear and Quadratic Inequalities

Solve linear inequalities, then quadratic ones using critical values and a sketch, write the answer correctly (and, or, set notation), and combine two conditions.

Cambridge 9709 Paper 1Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Solve linear inequalities, reversing the sign when multiplying or dividing by a negative.
Solve quadratic inequalities by finding the critical values and sketching.
Write solutions correctly as $a<x<b$ , as " $x<a$ or $x>b$ ", or in set notation.
Find the values that satisfy two inequalities at once, and list integer solutions.

Pause and think

Is it true that $x^2>9$ means $x>3$ ? Try $x=-4$ .

Section 2 of 14

Core idea

Start hereWhen does the business make money?

A small shop's daily profit (in hundreds of dollars) from selling $x$ hundred items is $P=-x^2+10x-16$ . It loses money if it sells too few (fixed costs) or too many (it has to slash prices). So the question isn't "what is $x$ ?" but "for which range of $x$ is $P>0$ ?" That's an inequality, and its answer is a whole interval of numbers. Here it turns out to be $2<x<8$ .

The big idea

To solve a quadratic inequality: (1) rearrange to compare with $0$ ; (2) find the critical values by solving the equation $=0$ ; (3) sketch the parabola; (4) read off where the graph is above ( $>0$ ) or below ( $<0$ ) the $x$ -axis. A ∪-shaped curve is negative between its roots and positive outside them.

1. The rules for inequalities

$<$ "less than", $>$ "greater than", $\le$ "less than or equal to", $\ge$ "greater than or equal to".
You may add or subtract the same number on both sides.
You may multiply or divide both sides by a positive number.
Multiplying or dividing by a negative number reverses the inequality.

Why a negative multiplier flips the sign

$a<b$
Start with any true inequality.
$a-a-b<b-a-b$
Subtract $a+b$ from both sides (allowed).
$-b<-a$
Simplify.
$-a>-b$
Read it right to left.

Multiplying by

-1

reverses the order:

2<5

but

-2>-5

Worked exampleLinear: $5-3x\le11$

Solve $5-3x\le11$ .

Goal: The set of $x$ .
Strategy: Isolate $x$ as for an equation, flipping when dividing by a negative.

$-3x\le6$
Subtract $5$ .
$x\ge-2$
Divide by $-3$ : flip $\le$ to $\ge$ .

Answer

x\ge-2

✓ CheckTry

x=0

(should work):

5\le11

✓. Try

x=-3

(shouldn't):

14\le11

✗ ✓.

2. Quadratic inequalities: the method

y = x² − x − 6: below the axis between −2 and 3

Shaded: the part of the curve below the axis, so

x^2-x-6<0

for

-2<x<3

. For

x^2-x-6>0

you would take the two outside parts:

x<-2

x>3

The answers on a number line

Open circles mean the end values are not included (strict

<

>

). For

\le

\ge

the circles are filled in and the end values are included.

Worked example $x^2-x-6<0$

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

Goal: All $x$ making the expression negative.
Strategy: Critical values, sketch, read off.

$x^2-x-6=0\Rightarrow(x-3)(x+2)=0\Rightarrow x=-2,\ 3$
Critical values: where it equals zero.
Sketch $y=x^2-x-6$ : ∪ shape ( $a=1>0$ ) crossing at $-2$ and $3$ .
" $<0$ " means "below the axis": between the roots.

Answer

-2<x<3

✓ CheckTest

x=0

(inside):

-6<0

✓. Test

x=4

(outside):

6<0

✗ ✓.

Worked exampleThe shop: $-x^2+10x-16>0$

For which $x$ is the profit $P=-x^2+10x-16$ positive?

Goal: Range of $x$ with $P>0$ .
Strategy: Multiply by $-1$ to make the $x^2$ coefficient positive (flip!), then use the method.

$x^2-10x+16<0$
Multiply by $-1$ : $>$ becomes $<$ .
$(x-2)(x-8)<0$
Product $16$ , sum $-10$ .
∪ shape, negative between the roots.

Answer

2<x<8

: sell between 200 and 800 items.

✓ Check

x=5

P=-25+50-16=9>0

✓.

x=1

P=-7<0

✓.

Worked exampleOutside the roots: $2x^2+x\ge3$

Solve $2x^2+x\ge3$ , giving the answer in set notation.

Goal: Set of $x$ .
Strategy: Rearrange to $\ge0$ ; " $\ge0$ " means on or above the axis: outside the roots, including them.

$2x^2+x-3\ge0$
Subtract $3$ .
$(2x+3)(x-1)\ge0$
$ac=-6$ : $3$ and $-2$ ; split and factorise.
Critical values $-\frac32$ and $1$ ; ∪ shape is $\ge0$ outside.

Answer

x\le-\frac32

x\ge1

; in set notation

\{x:x\le-\tfrac32\}\cup\{x:x\ge1\}

✓ Check

x=0

0\ge3

✗ (correctly excluded);

x=2

10\ge3

✓.

Common misconceptionWriting $3<x<-2$ "Outside the roots" is two separate pieces:

x<-2

x>3

. A single double inequality like

3<x<-2

is impossible (no number is both

>3

and

<-2

3. Special cases

Solving x² < k

k = 1.0

The bold stretch of the

x

-axis is the solution. Where the curve is below the dashed line

y=k

, the inequality

x^2<k

is true.

x^2<9

$x^2-9<0$
Subtract $9$ .
$(x-3)(x+3)<0$
Difference of two squares.
$-3<x<3$
Between the roots.

Not just

x<3

x=-5

gives

25<9

, false.

Edge caseNo critical values

x^2+2x+5>0

\Delta=4-20<0

, no roots; ∪ shape so it is always positive. Answer: all real

x

. And

x^2+2x+5<0

has no solutions.

Common misconceptionMultiplying by something that might be negative To solve

\frac{3}{x}<1

you can't multiply by

x

(it may be negative). Multiply by

x^2

instead (always positive):

3x<x^2

Solving a quadratic inequality

Drawing diagram…

Use $\le$ / $\ge$ to include the critical values themselves.

Pause and think

Solve $x^2\ge4x$ .

Summary

Flip the sign when multiplying or dividing by a negative.
Critical values → sketch → read.
∪ shape: $<0$ between, $>0$ outside.
Watch for no roots, and never multiply by an unknown sign.

Critical values: Where the expression equals zero; the boundaries.
Set notation: e.g. $\{x:x>2\}$ , "the set of $x$ such that $x>2$ ".

HistoryWho invented < and >? The symbols

<

and

>

first appeared in print in 1631, in a book by the English mathematician Thomas Harriot, published ten years after his death. The versions with a line underneath,

\le

and

\ge

, came about a century later, from the French scientist Pierre Bouguer in 1734.

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