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Algebra and functions · Subtopic 4 of 16

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<ba<x<b, as "x<ax<a or x>bx>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 x2>9x^2>9 means x>3x>3? Try x=−4x=-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 xx hundred items is P=−x2+10x−16P=-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 xx?" but "for which range of xx is P>0P>0?" That's an inequality, and its answer is a whole interval of numbers. Here it turns out to be 2<x<82<x<8.

The big idea

To solve a quadratic inequality: (1) rearrange to compare with 00; (2) find the critical values by solving the equation =0=0; (3) sketch the parabola; (4) read off where the graph is above (>0>0) or below (<0<0) the xx-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
  1. a<ba<b
    Start with any true inequality.
  2. a−a−b<b−a−ba-a-b<b-a-b
    Subtract a+ba+b from both sides (allowed).
  3. −b<−a-b<-a
    Simplify.
  4. −a>−b-a>-b
    Read it right to left.
Multiplying by −1-1 reverses the order: 2<52<5 but −2>−5-2>-5.
Worked exampleLinear: 5−3x≤115-3x\le11

Solve 5−3x≤115-3x\le11.

Goal
The set of xx.
Strategy
Isolate xx as for an equation, flipping when dividing by a negative.
  1. −3x≤6-3x\le6
    Subtract 55.
  2. x≥−2x\ge-2
    Divide by −3-3: flip ≤\le to ≥\ge.
Answerx≥−2x\ge-2.
✓ CheckTry x=0x=0 (should work): 5≤115\le11 ✓. Try x=−3x=-3 (shouldn't): 14≤1114\le11 ✗ ✓.

2. Quadratic inequalities: the method

y = x² − x − 6: below the axis between −2 and 3
−4−224−6−4−224680xyy = x² − x − 6y < 0 here−23
Shaded: the part of the curve below the axis, so x2−x−6<0x^2-x-6<0 for −2<x<3-2<x<3. For x2−x−6>0x^2-x-6>0 you would take the two outside parts: x<−2x<-2 or x>3x>3.
The answers on a number line
−23x² − x − 6 < 0: −2 < x < 3−23x² − x − 6 > 0: x < −2 or x > 3
Open circles mean the end values are not included (strict << or >>). For ≤\le or ≥\ge the circles are filled in and the end values are included.
Worked examplex2−x−6<0x^2-x-6<0

Solve x2−x−6<0x^2-x-6<0.

Goal
All xx making the expression negative.
Strategy
Critical values, sketch, read off.
  1. x2−x−6=0⇒(x−3)(x+2)=0⇒x=−2, 3x^2-x-6=0\Rightarrow(x-3)(x+2)=0\Rightarrow x=-2,\ 3
    Critical values: where it equals zero.
  2. Sketch y=x2−x−6y=x^2-x-6: ∪ shape (a=1>0a=1>0) crossing at −2-2 and 33.

  3. "<0<0" means "below the axis": between the roots.

Answer−2<x<3-2<x<3.
✓ CheckTest x=0x=0 (inside): −6<0-6<0 ✓. Test x=4x=4 (outside): 6<06<0 ✗ ✓.
Worked exampleThe shop: −x2+10x−16>0-x^2+10x-16>0

For which xx is the profit P=−x2+10x−16P=-x^2+10x-16 positive?

Goal
Range of xx with P>0P>0.
Strategy
Multiply by −1-1 to make the x2x^2 coefficient positive (flip!), then use the method.
  1. x2−10x+16<0x^2-10x+16<0
    Multiply by −1-1: >> becomes <<.
  2. (x−2)(x−8)<0(x-2)(x-8)<0
    Product 1616, sum −10-10.
  3. ∪ shape, negative between the roots.

Answer2<x<82<x<8: sell between 200 and 800 items.
✓ Checkx=5x=5: P=−25+50−16=9>0P=-25+50-16=9>0 ✓. x=1x=1: P=−7<0P=-7<0 ✓.
Worked exampleOutside the roots: 2x2+x≥32x^2+x\ge3

Solve 2x2+x≥32x^2+x\ge3, giving the answer in set notation.

Goal
Set of xx.
Strategy
Rearrange to ≥0\ge0; "≥0\ge0" means on or above the axis: outside the roots, including them.
  1. 2x2+x−3≥02x^2+x-3\ge0
    Subtract 33.
  2. (2x+3)(x−1)≥0(2x+3)(x-1)\ge0
    ac=−6ac=-6: 33 and −2-2; split and factorise.
  3. Critical values −32-\frac32 and 11; ∪ shape is ≥0\ge0 outside.

Answerx≤−32x\le-\frac32 or x≥1x\ge1; in set notation {x:x≤−32}∪{x:x≥1}\{x:x\le-\tfrac32\}\cup\{x:x\ge1\}.
✓ Checkx=0x=0: 0≥30\ge3 ✗ (correctly excluded); x=2x=2: 10≥310\ge3 ✓.
Common misconceptionWriting 3<x<−23<x<-2 "Outside the roots" is two separate pieces: x<−2x<-2 or x>3x>3. A single double inequality like 3<x<−23<x<-2 is impossible (no number is both >3>3 and <−2<-2).

3. Special cases

Solving x² < k
−4−3−2−112342468100xyy = x²
The bold stretch of the xx-axis is the solution. Where the curve is below the dashed line y=ky=k, the inequality x2<kx^2<k is true.
x2<9x^2<9
  1. x2−9<0x^2-9<0
    Subtract 99.
  2. (x−3)(x+3)<0(x-3)(x+3)<0
    Difference of two squares.
  3. −3<x<3-3<x<3
    Between the roots.
Not just x<3x<3: x=−5x=-5 gives 25<925<9, false.
Edge caseNo critical values x2+2x+5>0x^2+2x+5>0: Δ=4−20<0\Delta=4-20<0, no roots; ∪ shape so it is always positive. Answer: all real xx. And x2+2x+5<0x^2+2x+5<0 has no solutions.
Common misconceptionMultiplying by something that might be negative To solve 3x<1\frac{3}{x}<1 you can't multiply by xx (it may be negative). Multiply by x2x^2 instead (always positive): 3x<x23x<x^2.
Solving a quadratic inequality

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

Pause and think

Solve x2≥4xx^2\ge4x.

Summary

  • Flip the sign when multiplying or dividing by a negative.
  • Critical values → sketch → read.
  • ∪ shape: <0<0 between, >0>0 outside.
  • Watch for no roots, and never multiply by an unknown sign.
Key words
Critical values
Where the expression equals zero; the boundaries.
Set notation
e.g. {x:x>2}\{x:x>2\}, "the set of xx such that x>2x>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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