Quadratic and Non-linear Inequalities — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers inequalities involving a quadratic expression, such as x² < 9 or x² − 5x + 6 ≥ 0. By the end of this guide you should be able to find the critical values of a quadratic inequality, decide which region of the number line satisfies it, and write the answer correctly in inequality notation.

You should also be able to handle the simplest form x² < k and x² > k without factorising, list the integers in a solution set, check a region with a test value, and recognise the two cases where the usual pattern is reversed.

The organising idea is short and worth memorising: the critical values are not the answer — they are the boundaries. Solving the equation only tells you where the expression changes sign. A second decision, made from a sketch or a test value, tells you which side of those boundaries you actually want. Almost every mark lost in this topic comes from stopping after the first decision.

Key terms and definitions

Quadratic inequality — an inequality containing a squared term, for example x² − 5x + 6 > 0.

Critical values — the solutions of the corresponding equation, found by setting the expression equal to zero. They are the x-values where the curve crosses the x-axis.

Parabola — the curve of a quadratic. With a positive x² term it is U-shaped; with a negative x² term it is ∩-shaped.

Region — a stretch of the number line between or beyond the critical values, on which the expression keeps the same sign throughout.

Solution set — all the values of x that satisfy the inequality, written in inequality notation.

Strict and non-strict — < and > exclude the boundary; ≤ and ≥ include it.

Core concepts

Why there are always two steps

A quadratic expression is positive on some stretches of the number line and negative on others, and it can only change sign where it equals zero. That is what makes the critical values useful: they cut the number line into regions, and inside each region the sign never changes.

So the method is always the same. Find where the expression is zero. Then decide, region by region, where it is positive and where it is negative. Only then do you know which region answers the question.

The simplest form: x² compared with a number

For x² < 9, think about which numbers square to less than 9. Both 2 and −2 do; both 4 and −4 do not. The solution is everything between the boundaries: −3 < x < 3.

For x² > 9, it is everything outside: x < −3 or x > 3.

The trap here is forgetting the negative branch. Writing x < 3 as the whole answer to x² < 9 omits the lower boundary and admits x = −10, whose square is 100. Every x² inequality has two boundaries, ±√k, because both a number and its negative give the same square.

Note also that x² > −4 is true for every value of x, since a square is never negative, and x² < −4 has no solutions at all. These appear occasionally and cannot be solved by the standard method.

Finding the critical values by factorising

For a full quadratic, set the expression equal to zero and solve it exactly as you would any quadratic equation.

For x² − 5x + 6 > 0: solve x² − 5x + 6 = 0, which factorises to (x − 2)(x − 3) = 0, giving critical values x = 2 and x = 3.

If the quadratic does not factorise neatly, the quadratic formula gives the critical values just as well; nothing else about the method changes.

Rearrange first if necessary. An inequality such as x² < 4x − 3 must become x² − 4x + 3 < 0 before factorising, because the regions are defined by where the expression is zero, and that means having everything on one side.

Choosing the region

With a positive x² term the parabola is U-shaped, dipping below the axis between the roots and rising above it on either side. So:

• > 0 or ≥ 0 — the curve is above the axis outside the roots: x < 2 or x > 3.
• < 0 or ≤ 0 — the curve is below the axis between the roots: 2 < x < 3.

A quick sketch is the fastest reliable way to see this. It does not need to be accurate: two crossing points on a line and a U through them is enough to show which parts of the curve sit above the axis.

A negative x² term flips the picture, so check the sign of the squared term before applying the pattern. Multiplying through by −1 to make it positive is often easier — but doing so reverses the inequality sign, so −x² + x + 6 > 0 becomes x² − x − 6 < 0.

Writing the answer

The shape of the answer differs between the two cases, and the notation must match.

A "between" answer is a single inequality with x in the middle: 2 < x < 3.

An "outside" answer is two separate inequalities joined by the word "or": x < 2 or x > 3. It cannot be written as a single chain — 3 < x < 2 describes nothing at all, since no number is both greater than 3 and less than 2.

Carry the strictness across from the original. If the question used ≤, the boundaries are included and the answer uses ≤ too.

The test-value check

Pick any convenient number inside a region and substitute it into the original inequality. If the statement is true, that whole region belongs to the solution; if false, it does not.

For x² − 4 > 0, the critical values are ±2. Testing x = 0 in the middle region gives 0 − 4 = −4, which is not greater than 0, so the middle fails and the answer is the outside: x < −2 or x > 2.

Zero is usually the easiest test value, provided it is not one of the critical values.

Worked examples

Example 1: A "between" answer

Solve x² − 5x + 6 < 0.

Set the expression to zero: x² − 5x + 6 = 0 factorises as (x − 2)(x − 3) = 0, so the critical values are 2 and 3.

The x² term is positive, so the parabola is U-shaped and dips below the axis between the roots. The inequality asks for where the expression is negative, so the answer is 2 < x < 3.

Check with a test value: x = 2.5 gives 6.25 − 12.5 + 6 = −0.25, which is indeed less than 0. ✓

Example 2: An "outside" answer, with the boundaries included

Solve x² − x − 12 ≥ 0.

Factorise x² − x − 12 = 0 as (x − 4)(x + 3) = 0, giving critical values 4 and −3.

The parabola is U-shaped, so it is above the axis outside the roots. Because the inequality is ≥, the boundaries themselves are included.

The answer is x ≤ −3 or x ≥ 4.

Writing this as −3 ≥ x ≥ 4 would be meaningless, and writing it as a single chain is the commonest presentation error in this topic.

Example 3: Rearranging before solving

Solve x² + 2x < 8.

Bring everything to one side first: x² + 2x − 8 < 0.

Factorise: (x + 4)(x − 2) = 0, so the critical values are −4 and 2.

The expression is negative between the roots, so the answer is −4 < x < 2.

Test x = 0 in the original: 0 + 0 = 0, which is less than 8. ✓ Since 0 lies between −4 and 2, the region is confirmed.

Common mistakes and how to avoid them

Giving the critical values as the answer. They are boundaries. The answer is a range, and the question is not finished until you have said which range.

Losing the negative branch of x² < k. The boundaries are ±√k. Every such inequality has two.

Writing an "outside" answer as a single chain. Use two inequalities joined by "or".

Forgetting to rearrange. Everything must be on one side, compared with zero, before the critical values mean anything.

Not reversing the sign when multiplying by −1. Multiplying or dividing an inequality by a negative number flips <, >, ≤ and ≥.

Dropping the "equal to". If the original used ≤ or ≥, so must the answer.

Assuming "greater than always means outside". That holds for a U-shaped parabola. Check the sign of the x² term first.

Exam technique for "Quadratic and Non-linear Inequalities"

Write the equation version as an explicit first line — "x² − 5x + 6 = 0" — before solving it. Examiners award a method mark for reaching the correct critical values, and that mark is available even when the region chosen afterwards is wrong.

Sketch the parabola, however roughly. Two crossings and a U is enough, and it removes the guesswork from choosing the region.

Substitute a test value and write the substitution down. It is a complete check on the region and takes one line.

Read the inequality symbol twice: once when finding the region and once when writing the final answer, so the strictness matches.

If the question asks for integer values rather than a range, solve for the range first and then list the integers inside it. Listing without the range risks missing one at each end.

Quick revision summary

Solve the matching equation first: the solutions are the critical values, and they are the boundaries of the answer, not the answer itself.

Rearrange so that everything is on one side and compared with zero before factorising.

For a positive x² term: the expression is negative between the roots and positive outside them. So < 0 gives a single inequality such as 2 < x < 3, and > 0 gives two, written as x < 2 or x > 3.

For x² < k the answer is −√k < x < √k; for x² > k it is x < −√k or x > √k. Both branches always exist.

Multiplying an inequality by a negative number reverses the sign.

Match the strictness of the original: < and > exclude the boundaries, ≤ and ≥ include them.

Confirm the region with a test value — usually x = 0 — substituted into the original inequality.