Pure · Algebra & Functions · Year 1

Inequalities

An inequality's answer is a region, not a point. Master the one sign-flip rule and let the parabola's sketch reveal exactly which range you want.

9 min read 3 worked examples

An inequality asks a richer question than an equation. An equation says "where is this exactly equal?"; an inequality says "where is this bigger (or smaller)?" The answer is rarely a single value — it's a whole range of values. Solving $2x+1=7$ gives the one point $x=3$; solving $2x+1>7$ gives every point beyond it, $x>3$. We're no longer finding a spot on the number line, we're finding a region of it.

For linear inequalities, you solve almost exactly as you would an equation — with one crucial exception that comes straight from what "bigger" means. For quadratic inequalities, the sketch of the parabola becomes your most powerful tool.

Why the sign flips when you multiply by a negative

Here's the one rule everyone half-remembers — now here's why. Start with a true statement: $3 < 5$. Multiply both sides by $-1$: is $-3 < -5$? No! $-3$ is the bigger number. The correct statement is $-3 > -5$. Multiplying (or dividing) by a negative reverses the order of the number line — what was smaller is now larger. That's not a rule to memorise; it's a fact about how negatives flip the line. Whenever you multiply or divide an inequality by a negative, the sign must flip.

Quadratic inequalities: let the parabola decide

To solve something like $x^2 - x - 6 > 0$, the reliable method is always the same three moves:

1. Rearrange so one side is zero, then find the roots of the corresponding equation (where the curve crosses the $x$-axis).
2. Sketch the parabola — just a rough U (or $\cap$) through those roots.
3. Read off the region: for $>0$ you want where the curve is above the axis; for $<0$, below it.

Strategy — the sketch is not optional

Never try to solve a quadratic inequality in your head from the roots alone — that's how the two most common errors creep in. A ten-second sketch showing the parabola and its two roots makes the answer unambiguous: you can literally see which region is above or below the axis. Mark the roots, note whether the U opens up or down, and shade.

Cartesian axes with the parabola y equals x squared minus x minus 6, roots marked at -2 and 3, and the two outer regions where the curve sits above the x-axis shaded.
A sketch of $y=x^2-x-6$, with the roots marked and the region where the curve sits above the x-axis shaded — the solution to $x^2-x-6>0$.

Worked example 1

A linear inequality with a sign flip
Solve $4 - 3x \geq 19$.
MethodIsolate x, flipping the sign when dividing by a negative
1
Subtract $4$ from both sides: $-3x \geq 15$.
Subtracting doesn't affect the inequality direction.
2
Divide both sides by $-3$. Because we're dividing by a negative, reverse the sign: $x \leq -5$.
The crucial flip: $\geq$ becomes $\leq$.
3
So the solution is $x \leq -5$.
A whole range, not a single value.
4
Check with a value in the range, say $x=-6$: $4-3(-6)=4+18=22 \geq 19$ ✓.
Confirms the direction is right.
▪ solved
Examiner's eye

The single most-checked point in the whole topic is that flip in step 2. If you dislike dividing by a negative, there's a neat dodge: add $3x$ to both sides first to get $4 \geq 19 + 3x$, then $-15 \geq 3x$, so $x \leq -5$ — same answer, no flip needed, because you never divided by a negative.

Worked example 2

A quadratic inequality · reading the region off a sketch
Solve $x^2 - x - 6 > 0$.
MethodFind the roots, sketch, then read the region
1
It's already $>0$ on one side. Find the roots of $x^2-x-6=0$: factorise to $(x-3)(x+2)=0$, so $x=3$ and $x=-2$.
Roots are where the parabola meets the axis.
2
Sketch: the $x^2$ coefficient is positive, so it's a U-shape crossing the axis at $-2$ and $3$.
A rough sketch is enough.
3
We want where the curve is above the axis ($>0$). A U-shape is above the axis outside its two roots.
The U dips below only between the roots.
4
So the solution is $x < -2$ or $x > 3$.
Two separate regions, joined by 'or'.
▪ solved
The key insight

The shape of the answer follows the shape of the parabola. For a U-shape (positive $x^2$): "$>0$" gives the outside region ($x<-2$ or $x>3$), while "$<0$" would give the inside ($-2<x<3$, a single joined-up range). The words "and"/"or" aren't arbitrary — they mirror whether the region is one connected piece or two separated ones. Always let the sketch, not memory, decide.

Worked example 3

Combining conditions · a "between" answer in set notation
Find the set of values of $x$ for which $2x^2 + 5x - 3 < 0$, giving your answer in set notation.

A "$<0$" quadratic with an upward parabola gives the region between the roots — a single connected range.

MethodFactorise, sketch, then express the inner region
1
Factorise $2x^2+5x-3 = (2x-1)(x+3)$, so the roots are $x=\tfrac12$ and $x=-3$.
Set each factor to zero.
2
The $x^2$ coefficient ($2$) is positive: a U-shape crossing at $-3$ and $\tfrac12$.
Sketch the parabola.
3
We want where the curve is below the axis ($<0$) — that's the dip between the roots.
Inside the U, between the crossings.
4
So $-3 < x < \tfrac12$.
A single connected range.
5
In set notation: $\{x : -3 < x < \tfrac12\}$.
The set of all x strictly between the roots.
▪ solved
Look deeper — strict vs inclusive

Because the inequality was strict ($<$, not $\leq$), the roots themselves are excluded — at $x=-3$ and $x=\tfrac12$ the expression equals $0$, not less than it. That's why we use $<$ (open) rather than $\leq$ (closed) in the answer, and why set notation uses strict inequalities here. Getting the boundary type right is a genuine mark.

Common mistake

Writing a quadratic answer as $3 < x < -2$ is meaningless — the left number must be smaller than the right. And never glue two outside regions with "and": $x<-2$ and $x>3$ describes no numbers at all. Use "or" for two separated regions, and always put the smaller bound on the left.

Check yourself

Solve the inequality $x^2 + 2x - 8 \leq 0$, giving your answer as a single range.
(Hint: factorise, sketch the U, and decide whether you want inside or outside.)

Method
1
Factorise: $x^2+2x-8=(x+4)(x-2)$, roots at $x=-4$ and $x=2$.
2
Positive $x^2$, so a U-shape crossing at $-4$ and $2$.
3
We want $\leq 0$: the region on or below the axis — that's between the roots (inclusive).
The dip inside the U.
4
So $-4 \leq x \leq 2$.
Inclusive, because of the $\leq$.
▪ solved

Because the inequality is inclusive ($\leq$), the roots are part of the solution, so we use closed bounds $-4 \leq x \leq 2$ — the single connected range where the U sits on or below the axis.