Mathematics 9709 · Pure Mathematics 1

Quadratic inequalities

A quadratic inequality asks which $x$ make a quadratic positive or negative.
The whole method: find the roots, sketch the parabola, read off the interval.

The one idea

Find the roots. Then read the side you want (above or below the axis) from a small sketch.

A parabola crosses the $x$ -axis at its roots. Between and outside the roots, it sits above (positive) or below (negative) the axis.
The sign of the $x^2$ term tells you which way it opens.

$x^2$	Shape	Below axis	Above axis
positive	smile $\smile$ (up)	between	outside
negative	frown $\frown$ (down)	outside	between

Here is the key fact for a smile: below the axis is between the roots, and above the axis is outside them.

An upward parabola: below the axis between the roots, above it outside them.

The three moves

Every quadratic inequality uses the same three steps: factorise → decide the shape → read the side.

Get one side to $0$ , then find the roots (calculator, §1.1, or factorise).
Decide the shape from the sign of $x^2$ (smile or frown).
Pick the side you want (above or below) and read the interval off the sketch.
Match endpoints: $<,>$ exclude the roots; $\le,\ge$ include them.

Starting from zero: finding roots, and the simplest case

It all starts with the roots, so practise that first. No inequality yet. To factorise $x^2-x-6$ , find two numbers that multiply to $-6$ and add to $-1$ . Those are $-3$ and $+2$ , since $(-3)\times(+2)=-6$ and $(-3)+(+2)=-1$ .

$x^2-x-6=(x-3)(x+2)$

The roots are the $x$ that make each bracket zero. They are $x=3$ and $x=-2$ . Those two numbers are all the sketch needs.

The simplest inequality is a difference of two squares. For $x^2-4<0$ , factorise to $(x-2)(x+2)<0$ , roots $x=\pm 2$ . The $x^2$ term is positive, so it is a smile. We want it below the axis, which is the middle part:

$-2<x<2$

Two worked solves

Is the $x^2$ term negative? Multiply every term by $-1$ first, and flip the inequality sign.

Worked example

(standard) Solve $x^2-x-6\ge 0$

Already $=0$ on one side. Factorise: $(x-3)(x+2)\ge 0$ , roots $x=3$ and $x=-2$ .
Positive $x^2$ → smile crossing at $-2$ and $3$ .
Want $\ge 0$ (on or above) → the two arms, outside the roots, endpoints included.

Answer

x\le -2\quad\text{or}\quad x\ge 3

Smile, roots −2 and 3; ≥ 0 wants on or above the axis, so the gold arms outside the roots.

Worked example

(exam level) Solve $12-x-x^2\ge 0$

Here the $x^2$ term is negative, so make it positive first.

Multiply by $-1$ and flip the sign: $x^2+x-12\le 0$ .
Factorise: $(x+4)(x-3)\le 0$ , roots $x=-4$ and $x=3$ .
Smile, want $\le 0$ (below) → between the roots, endpoints included.

Answer

-4\le x\le 3

The frown y = 12 − x − x²; ≥ 0 is the part on or above the axis, the gold band between the roots −4 and 3.

Why multiplying by −1 flips the sign

Multiplying an inequality by a negative number reverses it, so $\ge$ becomes $\le$ :

$-12+x+x^2\le 0\ \Longrightarrow\ x^2+x-12\le 0$

And the factorising: two numbers multiplying to $-12$ , adding to $+1$ are $+4$ and $-3$ .

If flipping feels risky, you can sketch the original frown and read the same interval. Most people find the “make it a smile” way safer in an exam.

Your turn— tap to reveal the worked answer (quick check)

Solve $x^2+2x-15<0$ . (9709-style)

Factorise: two numbers multiplying to $-15$ , adding to $+2$ → $+5$ and $-3$ , so $(x+5)(x-3)<0$ ; roots $-5$ and $3$ . Smile, want $<0$ (below) → between them.

Answer

-5<x<3

Where it shows up

Usually the inequality is the last step of a §1.5 discriminant question. Finish it in exactly this way.

A plain “solve this inequality” on its own is quite rare. Far more often you work out a condition like $k^2-2k-15>0$ and have to finish it. (9709/12 Mar 2021 Q4)

Answer

k<-3\quad\text{or}\quad k>5

The full working

Two numbers multiplying to $-15$ , adding to $-2$ are $-5$ and $+3$ , giving $(k-5)(k+3)>0$ with roots $-3$ and $5$ . It's an upward parabola in $k$ and we want $>0$ (strictly above the axis), so outside the roots, endpoints excluded.

So this skill is really the end of the discriminant topic. It is worth practising until it is easy.

Where the marks go

Common mistake

Never take the square root of an inequality.

x^2<9

-3<x<3

, not

x<3

. Always go through the roots and a sketch.

Common mistake

The answer is an interval (or two), not a single value. “Outside the roots” is two pieces joined by or:

x<-2

x>3

. Never write it as one chain like

3<x<-2

, which means nothing. And multiplying by a negative flips the sign. Forgetting this is the common frown mistake.

Now you try

One more, with a leading coefficient that isn't $1$ .

Solve $2x^2+x-6\ge 0$ . (9709-style)

Your turn— tap to reveal the worked answer (leading coefficient isn't 1, so factorise carefully)

Factorise $2x^2+x-6$ with the split-the-middle method from Solving by factorising. Find two numbers that multiply to $2\times(-6)=-12$ and add to $+1$ . They are $+4$ and $-3$ . Split the middle term and group:

$2x^2+4x-3x-6 = 2x(x+2)-3(x+2) = (2x-3)(x+2)$

Roots: $2x-3=0\Rightarrow x=\tfrac{3}{2}$ and $x+2=0\Rightarrow x=-2$ .

Positive $x^2$ term → smile. We want $\ge 0$ (on or above) → outside the roots, endpoints included.

Answer

x\le -2\quad\text{or}\quad x\ge \tfrac{3}{2}

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Quadratic inequalities

A quadratic inequality asks which $x$ make a quadratic positive or negative.
The whole method: find the roots, sketch the parabola, read off the interval.

The one idea

Find the roots. Then read the side you want (above or below the axis) from a small sketch.

A parabola crosses the $x$ -axis at its roots. Between and outside the roots, it sits above (positive) or below (negative) the axis.
The sign of the $x^2$ term tells you which way it opens.

$x^2$	Shape	Below axis	Above axis
positive	smile $\smile$ (up)	between	outside
negative	frown $\frown$ (down)	outside	between

Here is the key fact for a smile: below the axis is between the roots, and above the axis is outside them.

An upward parabola: below the axis between the roots, above it outside them.

The three moves

Every quadratic inequality uses the same three steps: factorise → decide the shape → read the side.

Get one side to $0$ , then find the roots (calculator, §1.1, or factorise).
Decide the shape from the sign of $x^2$ (smile or frown).
Pick the side you want (above or below) and read the interval off the sketch.
Match endpoints: $<,>$ exclude the roots; $\le,\ge$ include them.

Starting from zero: finding roots, and the simplest case

$x^2-x-6=(x-3)(x+2)$

The roots are the $x$ that make each bracket zero. They are $x=3$ and $x=-2$ . Those two numbers are all the sketch needs.

$-2<x<2$

Two worked solves

Is the $x^2$ term negative? Multiply every term by $-1$ first, and flip the inequality sign.

Worked example

(standard) Solve $x^2-x-6\ge 0$

Already $=0$ on one side. Factorise: $(x-3)(x+2)\ge 0$ , roots $x=3$ and $x=-2$ .
Positive $x^2$ → smile crossing at $-2$ and $3$ .
Want $\ge 0$ (on or above) → the two arms, outside the roots, endpoints included.

Answer

x\le -2\quad\text{or}\quad x\ge 3

Smile, roots −2 and 3; ≥ 0 wants on or above the axis, so the gold arms outside the roots.

Worked example

(exam level) Solve $12-x-x^2\ge 0$

Here the $x^2$ term is negative, so make it positive first.

Multiply by $-1$ and flip the sign: $x^2+x-12\le 0$ .
Factorise: $(x+4)(x-3)\le 0$ , roots $x=-4$ and $x=3$ .
Smile, want $\le 0$ (below) → between the roots, endpoints included.

Answer

-4\le x\le 3

The frown y = 12 − x − x²; ≥ 0 is the part on or above the axis, the gold band between the roots −4 and 3.

Why multiplying by −1 flips the sign

Multiplying an inequality by a negative number reverses it, so $\ge$ becomes $\le$ :

$-12+x+x^2\le 0\ \Longrightarrow\ x^2+x-12\le 0$

And the factorising: two numbers multiplying to $-12$ , adding to $+1$ are $+4$ and $-3$ .

If flipping feels risky, you can sketch the original frown and read the same interval. Most people find the “make it a smile” way safer in an exam.

Your turn— tap to reveal the worked answer (quick check)

Solve $x^2+2x-15<0$ . (9709-style)

Factorise: two numbers multiplying to $-15$ , adding to $+2$ → $+5$ and $-3$ , so $(x+5)(x-3)<0$ ; roots $-5$ and $3$ . Smile, want $<0$ (below) → between them.

Answer

-5<x<3

Where it shows up

Usually the inequality is the last step of a §1.5 discriminant question. Finish it in exactly this way.

A plain “solve this inequality” on its own is quite rare. Far more often you work out a condition like $k^2-2k-15>0$ and have to finish it. (9709/12 Mar 2021 Q4)

Answer

k<-3\quad\text{or}\quad k>5

The full working

So this skill is really the end of the discriminant topic. It is worth practising until it is easy.

Where the marks go

Common mistake

Never take the square root of an inequality.

x^2<9

-3<x<3

, not

x<3

. Always go through the roots and a sketch.

Common mistake

The answer is an interval (or two), not a single value. “Outside the roots” is two pieces joined by or:

x<-2

x>3

. Never write it as one chain like

3<x<-2

, which means nothing. And multiplying by a negative flips the sign. Forgetting this is the common frown mistake.

Now you try

One more, with a leading coefficient that isn't $1$ .

Solve $2x^2+x-6\ge 0$ . (9709-style)

Your turn— tap to reveal the worked answer (leading coefficient isn't 1, so factorise carefully)

$2x^2+4x-3x-6 = 2x(x+2)-3(x+2) = (2x-3)(x+2)$

Roots: $2x-3=0\Rightarrow x=\tfrac{3}{2}$ and $x+2=0\Rightarrow x=-2$ .

Positive $x^2$ term → smile. We want $\ge 0$ (on or above) → outside the roots, endpoints included.

Answer

x\le -2\quad\text{or}\quad x\ge \tfrac{3}{2}