1MA1 · A22AlgebraHigherGrades 7–8

Quadratic Inequalities

Q: Why do I need a sketch?

It shows at a glance which region is above or below the axis, which is much more reliable than trying to reason about signs.

Q: Can I square root both sides of x^2 < 25?

Not safely — it gives x < 5 and misses the lower limit. The solution is -5 < x < 5.

Q: What if the quadratic has no roots?

A U-shaped curve that never meets the axis is positive for all x, so "> 0" is true for every x and "< 0" has no solutions.

Solving $x^{2} - x - 12 < 0$ is not like solving a linear inequality. You cannot just rearrange for $x$ , because the expression is negative on one stretch of the number line and positive on two others. The reliable method combines two things you already know: solve the quadratic equation to find the boundaries, then sketch the parabola to see which side you need.

45 min Quadratics

The big picture

The roots of the quadratic — the $\textbf{critical values}$ — split the number line into regions. On a U-shaped parabola, the expression is negative between the roots and positive outside them. The sketch makes that visible, so you never have to guess.

Edexcel Higher questions ask for the solution as an inequality, in set notation, or as a list of integers, and sometimes combine a quadratic and a linear inequality. Answers must describe two separate intervals correctly when the solution lies outside the roots.

What you'll learn

Find critical values by solving the corresponding quadratic equation
Use a sketch to solve quadratic inequalities
Write solutions as inequalities and in set notation
Find integer solutions and combine with linear inequalities

The method

Step 1: rearrange so one side is 0, with a positive $x^{2}$ coefficient if possible.

Step 2: solve the equation $ax^{2} + bx + c = 0$ to find the critical values.

Step 3: sketch the parabola through those roots.

Step 4: for " $< 0$ " take where the curve is below the $x$ -axis; for " $> 0$ " take where it is above.

Example 1Solve

x^{2} - x - 12 < 0

1Critical values:

(x - 4)(x + 3) = 0

, so

x = -3

and

x = 4

2U-shaped curve through

-3

and

4

3Below the axis between the roots.

Answer

-3 < x < 4

Your turnSame method, new numbers — try it before moving on.

Solve $x^2 - 2x - 8 \le 0$ .

Your answer

Type it like -2 <= x <= 4.

Show worked solution

Solution

1 $(x - 4)(x + 2) = 0$ : critical values $-2$ and $4$ .
2U-shape below the axis between the roots.

Answer: $-2 \le x \le 4$

Why this works

Test a value to check: $x = 0$ lies between the roots and gives $-12 < 0$ ✓, while $x = 5$ lies outside and gives $8$ , which is not less than 0.

Solutions outside the roots

When the inequality is " $> 0$ " or " $\ge 0$ " on a U-shaped curve, the solution is in two parts: to the left of the smaller root $\textit{and}$ to the right of the larger root.

Write this as two separate inequalities joined by "or": $x \le -3$ or $x \ge 4$ . It cannot be written as one double inequality.

Example 1Solve

2x^{2} + x - 6 \ge 0

1Critical values:

(2x - 3)(x + 2) = 0

, so

x = -2

and

x = \tfrac32

2U-shaped: on or above the axis outside the roots, including the roots themselves.

Answer

x \le -2

x \ge \tfrac32

Your turnSame method, new numbers — try it before moving on.

Solve $x^2 + 3x - 10 > 0$ .

Your answer

Type it like x < -5 or x > 2.

Show worked solution

Solution

1 $(x + 5)(x - 2) = 0$ : critical values $-5$ and $2$ .
2Above the axis outside the roots.

Answer: $x < -5$ or $x > 2$

Tip — Writing $\tfrac32 \le x \le -2$ is impossible — no number is both at least $\tfrac32$ and at most $-2$ . Two regions always need "or".

Rearranging first

Terms on both sides must be gathered first: $x^{2} \ge 9$ becomes $x^{2} - 9 \ge 0$ , with critical values $\pm 3$ .

Do not square root both sides of an inequality — it loses the negative region.

If the $x^{2}$ coefficient is negative, multiply through by $-1$ and reverse the inequality sign, or sketch the ∩-shaped curve directly.

Example 1Solve

x^{2} \ge 9

x^{2} - 9 \ge 0

, so

(x + 3)(x - 3) \ge 0

2U-shaped, on or above the axis outside

\pm 3

Answer

x \le -3

x \ge 3

Set notation and integers

In set notation, $-3 < x < 4$ is $\{x : -3 < x < 4\}$ , and two regions are joined with the union symbol: $\{x : x \le -2\} \cup \{x : x \ge \tfrac32\}$ .

For integer solutions, list the whole numbers in the solution set, checking the endpoints.

Example 1Find the integers that satisfy both

x^{2} < 16

and

2x + 1 > 0

x^{2} < 16

gives

-4 < x < 4

2x + 1 > 0

gives

x > -\tfrac12

3Both:

-\tfrac12 < x < 4

Answer

x = 0, 1, 2, 3

Your turnSame method, new numbers — try it before moving on.

List the integers that satisfy both $x^2 \le 10$ and $x > 1$ .

Your answer

Separate with commas.

Show worked solution

Solution

1 $x^2 \le 10$ gives $-3.16 \le x \le 3.16$ .
2With $x > 1$ : $x = 2, 3$ .

Answer: $2, 3$

Think like an examiner

Show the critical values and a sketch — both earn method marks.
Use "or" (or $\cup$ ) for solutions in two parts.
Match strict and inclusive signs to the original inequality.

Quadratic inequalities

(x - a)(x - b) < 0 \Rightarrow a < x < b

Between the roots (a < b).

(x - a)(x - b) > 0 \Rightarrow x < a \text{ or } x > b

Outside the roots.

A \cup B

Union: combine two regions.

Watch out for these

Solving

x^{2} > 9

x > 3

only.

Include the negative region:

x < -3

x > 3

Writing

4 < x < -3

Order the values correctly:

-3 < x < 4

, or use "or" for outside regions.

Taking the region between the roots for "

> 0

For a U-shape, "

> 0

" is outside the roots.

Forgetting to rearrange

x^{2} < 2x + 8

first.

Write

x^{2} - 2x - 8 < 0

before finding critical values.

Stretch yourself

Find the set of values of $x$ for which $x^{2} - 2x - 8 < 0$ and $3x - 2 \ge 1$ are both true.

Hint — Solve each inequality separately, then find the overlap on a number line.

Reveal solution

SolutionFirst:

(x - 4)(x + 2) < 0

, so

-2 < x < 4

. Second:

3x \ge 3

, so

x \ge 1

. The overlap of

-2 < x < 4

and

x \ge 1

1 \le x < 4

, or in set notation

\{x : 1 \le x < 4\}

. Check

x = 1

1 - 2 - 8 = -9 < 0

✓ and

3 - 2 = 1 \ge 1

✓.

\blacksquare

Questions students ask

Why do I need a sketch?

It shows at a glance which region is above or below the axis, which is much more reliable than trying to reason about signs.

Can I square root both sides of

x^{2} < 25

Not safely — it gives

x < 5

and misses the lower limit. The solution is

-5 < x < 5

What if the quadratic has no roots?

A U-shaped curve that never meets the axis is positive for all

x

, so "

> 0

" is true for every

x

and "

< 0

" has no solutions.

Key takeaways

Rearrange to zero and find the critical values.
Sketch the parabola through the roots.
U-shape: negative between the roots, positive outside.
Two-part solutions use "or" or $\cup$ .
Combine with other inequalities by finding the overlap.

Exam-style questions

Practise it the way it is examined

2 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 73 marks

Solve $3x^2 - 7x - 6 < 0$ .[3]
Show mark scheme
- M1 $(3x + 2)(x - 3)$
- M1Critical values $-\frac{2}{3}$ and $3$
- A1 $-\frac{2}{3} < x < 3$
Answer: $-\frac{2}{3} < x < 3$

Question 2Grade 83 marks

Solve $x^2 > 4x + 12$ .[3]
Show mark scheme
- M1 $x^2 - 4x - 12 > 0$
- M1 $(x - 6)(x + 2)$
- A1 $x < -2$ or $x > 6$
Answer: $x < -2$ or $x > 6$

Unlimited practice

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