N8NumberHigherGrades 6–9

Surds

Q: Why rationalise at all?

Convention and convenience: an answer like 63 is easier to compare and combine than 2 6, and exam questions ask for this form.

A surd is a square root that does not simplify to a whole number, such as $\sqrt{2}$ or $3\sqrt{5}$ . Surds let you give answers exactly — a calculator can only ever show a rounded decimal. Higher papers use surds in Pythagoras, trigonometry, quadratic equations and "show that" proofs, so fluency here pays off across many topics.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

Two rules do almost all the work: $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ . They let you pull square factors out of a root, and multiply and divide surds freely.

Treat a surd like an algebraic letter. $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ just as $2x + 5x = 7x$ , and $(2 + \sqrt{3})(4 - \sqrt{3})$ expands exactly like $(2 + x)(4 - x)$ — except that $\sqrt{3} \times \sqrt{3} = 3$ .

SurdsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Simplify surds by finding square factors
Add, subtract, multiply and divide surds
Expand brackets containing surds
Rationalise the denominator of a fraction

Simplifying surds

Find the largest square number that is a factor of the number under the root. Split the root and take the square root of that factor.

Square numbers to look for: $4, 9, 16, 25, 36, 49, 64, 81, 100$ .

Example 1Simplify

\sqrt{72}

1The largest square factor of 72 is 36:

72 = 36 \times 2

\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}

Answer

6\sqrt{2}

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

Simplify $\sqrt{48}$ .

Your answer

Type √ as sqrt, e.g. 5sqrt2.

Show worked solution

Solution

1 $48 = 16 \times 3$ .
2 $\sqrt{48} = 4\sqrt{3}$ .

Answer: $4\sqrt{3}$

Tip — Using a smaller square factor still works but needs another step: $\sqrt{72} = 2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$ .

Calculating with surds

Only like surds can be added or subtracted — simplify first to see whether they match. $\sqrt{2} + \sqrt{3}$ cannot be simplified.

Multiply the whole numbers together and the roots together: $3\sqrt{2} \times 4\sqrt{5} = 12\sqrt{10}$ . Remember $\sqrt{a} \times \sqrt{a} = a$ .

Example 1Simplify

\sqrt{50} + \sqrt{18} - \sqrt{8}

\sqrt{50} = 5\sqrt{2}

\sqrt{18} = 3\sqrt{2}

\sqrt{8} = 2\sqrt{2}

5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2}

Answer

6\sqrt{2}

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

Simplify $\sqrt{75} - \sqrt{12}$ .

Your answer

Type √ as sqrt, e.g. 5sqrt2.

Show worked solution

Solution

1 $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{12} = 2\sqrt{3}$ .
2 $5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3}$ .

Answer: $3\sqrt{3}$

Example 2Simplify

2\sqrt{6} \times 5\sqrt{3}

2 \times 5 = 10

and

\sqrt{6} \times \sqrt{3} = \sqrt{18} = 3\sqrt{2}

10 \times 3\sqrt{2}

Answer

30\sqrt{2}

Expanding brackets

Expand exactly as in algebra, then simplify. Products such as $\sqrt{5} \times \sqrt{5}$ become whole numbers.

A useful special case: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ . The surd terms cancel, leaving a whole number.

Example 1Expand and simplify

(3 + \sqrt{2})(5 - \sqrt{2})

15 - 3\sqrt{2} + 5\sqrt{2} - 2

2Collect:

13 + 2\sqrt{2}

Answer

13 + 2\sqrt{2}

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

Expand and simplify $(4 - \sqrt{3})^2$ .

Your answer

Type it like 7 + 4sqrt3.

Show hint

Write it as $(4 - \sqrt{3})(4 - \sqrt{3})$ .

Show worked solution

Solution

1 $16 - 4\sqrt{3} - 4\sqrt{3} + 3$ .
2Collect: $19 - 8\sqrt{3}$ .

Answer: $19 - 8\sqrt{3}$

Rationalising the denominator

To remove a surd from a denominator, multiply the top and bottom by that surd: $\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$ .

For a denominator like $3 + \sqrt{2}$ , multiply top and bottom by $3 - \sqrt{2}$ — the difference of two squares clears the surd.

Example 1Rationalise the denominator of

\dfrac{12}{\sqrt{3}}

1Multiply by

\frac{\sqrt{3}}{\sqrt{3}}

\frac{12\sqrt{3}}{3}

2Simplify:

4\sqrt{3}

Answer

4\sqrt{3}

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

Rationalise the denominator of $\dfrac{10}{\sqrt{5}}$ .

Your answer

Type √ as sqrt, e.g. 3sqrt2.

Show worked solution

Solution

1 $\frac{10\sqrt{5}}{5}$ .
2 $= 2\sqrt{5}$ .

Answer: $2\sqrt{5}$

Example 2Rationalise the denominator of

\dfrac{4}{3 + \sqrt{5}}

1Multiply top and bottom by

3 - \sqrt{5}

2Denominator:

(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4

3Numerator:

4(3 - \sqrt{5})

. Divide by 4.

Answer

3 - \sqrt{5}

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

Rationalise the denominator of $\dfrac{6}{4 - \sqrt{10}}$ .

Your answer

Type it like 3 + sqrt5.

Show hint

Multiply by $\frac{4 + \sqrt{10}}{4 + \sqrt{10}}$ . The denominator becomes $16 - 10$ .

Show worked solution

Solution

1Denominator: $16 - 10 = 6$ .
2Numerator: $6(4 + \sqrt{10})$ .
3Divide by 6: $4 + \sqrt{10}$ .

Answer: $4 + \sqrt{10}$

Multiplying by 1

Rationalising does not change the value — you multiply by a fraction equal to 1. It simply puts the answer in a form that is easier to compare and to combine with other surds.

Think like an examiner

"Give your answer in the form $a\sqrt{b}$ " means $b$ must be as small as possible: $2\sqrt{18}$ is not accepted where $6\sqrt{2}$ is expected.
Show the multiplication when rationalising, including the expanded denominator.
Exact answers mean surds: do not convert to decimals unless asked.

Surd rules

\sqrt{ab} = \sqrt{a}\sqrt{b}, \quad \sqrt{\tfrac{a}{b}} = \tfrac{\sqrt{a}}{\sqrt{b}}

Split and combine roots.

\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}

Rationalising a single surd.

(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b

Clears the surd from a denominator.

Watch out for these

\sqrt{9 + 16} = 3 + 4

Roots do not split over addition:

\sqrt{25} = 5

\sqrt{2} + \sqrt{3} = \sqrt{5}

Unlike surds cannot be combined.

(\sqrt{5})^2 = 25

\sqrt{5} \times \sqrt{5} = 5

Rationalising

\frac{1}{2 + \sqrt{3}}

by multiplying by

\sqrt{3}

Multiply by

2 - \sqrt{3}

to use the difference of two squares.

Stretch yourself

A rectangle has sides $(3 + \sqrt{2})$ cm and $(3 - \sqrt{2})$ cm. Show that its diagonal is $\sqrt{22}$ cm.

Hint — Use Pythagoras. Square each side carefully.

Reveal solution

Solution

(3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}

and

(3 - \sqrt{2})^2 = 11 - 6\sqrt{2}

. Adding:

22

. So the diagonal is

\sqrt{22}

cm.

\blacksquare

Questions students ask

\sqrt{9}

a surd?

No — it equals 3 exactly. A surd is a root that is irrational, like

\sqrt{10}

Why rationalise at all?

Convention and convenience: an answer like

\frac{\sqrt{6}}{3}

is easier to compare and combine than

\frac{2}{\sqrt{6}}

, and exam questions ask for this form.

Can I use a calculator for surds?

Many calculators simplify surds, which is useful for checking. But surd questions are usually on the non-calculator paper, so learn the methods.

Key takeaways

Simplify by taking out the largest square factor.
Add and subtract only like surds.
Multiply whole numbers and roots separately; $\sqrt{a} \times \sqrt{a} = a$ .
Expand brackets with surds like algebra.
Rationalise: multiply by $\frac{\sqrt{b}}{\sqrt{b}}$ , or by the conjugate $a - \sqrt{b}$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 8 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 63 marks

(a)Simplify $\sqrt{200}$ .[1]
Show mark scheme (a)
- B1 $10\sqrt{2}$
Answer: $10\sqrt{2}$
(b)Show that $\sqrt{27} + \sqrt{48} = 7\sqrt{3}$ .[2]
Show mark scheme (b)
- M1Simplifies one surd correctly: $3\sqrt{3}$ or $4\sqrt{3}$
- A1 $3\sqrt{3} + 4\sqrt{3} = 7\sqrt{3}$ with both shown

Question 2Grade 72 marks

Expand and simplify $(5 + 2\sqrt{3})(1 - \sqrt{3})$ . Give your answer in the form $a + b\sqrt{3}$ .[2]
Show mark scheme
- M1Four terms with at least three correct: $5 - 5\sqrt{3} + 2\sqrt{3} - 6$
- A1 $-1 - 3\sqrt{3}$
Answer: $-1 - 3\sqrt{3}$

Question 3Grade 83 marks

Rationalise the denominator and simplify $\dfrac{2 + \sqrt{7}}{3 - \sqrt{7}}$ .[3]
Show mark scheme
- M1Multiplies numerator and denominator by $3 + \sqrt{7}$
- M1Denominator $9 - 7 = 2$ ; numerator expanded $6 + 2\sqrt{7} + 3\sqrt{7} + 7$
- A1 $\dfrac{13 + 5\sqrt{7}}{2}$
Answer: $\dfrac{13 + 5\sqrt{7}}{2}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Simplify $\sqrt{20}$ .
- $2\sqrt{5}$
- $4\sqrt{5}$
- $5\sqrt{2}$
- $10\sqrt{2}$
Show answer and explanation
Answer: $2\sqrt{5}$
$20 = 4 \times 5$ .
Simplify $\sqrt{3} \times \sqrt{12}$ .
- $6$
- $\sqrt{15}$
- $36$
- $3\sqrt{2}$
Show answer and explanation
Answer: $6$
$\sqrt{36} = 6$ .
Simplify $4\sqrt{5} - \sqrt{5}$ .
- $3\sqrt{5}$
- $4$
- $3$
- $\sqrt{15}$
Show answer and explanation
Answer: $3\sqrt{5}$
Like surds: $4 - 1 = 3$ .
Simplify $\sqrt{32} + \sqrt{2}$ .
- $5\sqrt{2}$
- $\sqrt{34}$
- $4\sqrt{2}$
- $17\sqrt{2}$
Show answer and explanation
Answer: $5\sqrt{2}$
$4\sqrt{2} + \sqrt{2}$ .
Rationalise $\frac{6}{\sqrt{2}}$ .
- $3\sqrt{2}$
- $6\sqrt{2}$
- $\frac{\sqrt{6}}{2}$
- $12$
Show answer and explanation
Answer: $3\sqrt{2}$
$\frac{6\sqrt{2}}{2}$ .
Expand $(\sqrt{3} + 1)(\sqrt{3} - 1)$ .
- $2$
- $4$
- $3 - \sqrt{3}$
- $2\sqrt{3}$
Show answer and explanation
Answer: $2$
$3 - 1$ .
Expand $(2 + \sqrt{5})^2$ .
- $9 + 4\sqrt{5}$
- $9$
- $4 + 5$
- $7 + 4\sqrt{5}$
Show answer and explanation
Answer: $9 + 4\sqrt{5}$
$4 + 4\sqrt{5} + 5$ .
Rationalise $\frac{1}{\sqrt{2} - 1}$ .
- $\sqrt{2} + 1$
- $\sqrt{2} - 1$
- $\frac{\sqrt{2}}{2}$
- $1$
Show answer and explanation
Answer: $\sqrt{2} + 1$
Denominator $(\sqrt{2} - 1)(\sqrt{2} + 1) = 1$ .

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 11 lessons