Surds · Study Notes · Edexcel IGCSE 4MA1 Higher

Manipulate roots of non-perfect squares without losing precision. The conjugate trick is one of Edexcel's favourite A* tools — guaranteed on Paper 1H.

At a glance

Surd = root of a non-perfect square (or higher power).
Simplify: pull out LARGEST perfect square factor.
Multiply: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ .
Square: $(\sqrt{a})^{2} = a$ .
Add/sub: only with SAME radical.
Rationalise: multiply by $\sqrt{c}$ or conjugate.
Conjugate: $a + b\sqrt{c} \leftrightarrow a - b\sqrt{c}$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.6 — Use surds and simplify expressions involving surds.
1.6 — Rationalise the denominator of a fraction.
1.6 — Use surds in geometric and algebraic contexts.

Simplify surds

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Pull out the LARGEST perfect-square factor.

Definition. A SURD is the root of a non-perfect square (or higher non-perfect power). $\sqrt{2}$ and $\sqrt[3]{5}$ are surds; $\sqrt{4} = 2$ is not.

Why surds matter. Many quantities (Pythagorean lengths, trig values, quadratic solutions) come out as surds. Edexcel mark schemes prefer EXACT answers in surd form rather than decimal approximations.

Method to simplify $\sqrt{n}$ :

Find the LARGEST perfect square factor of $n$ .
Use $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ .

Common perfect squares to know: $4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225$ .

Split off the largest perfect-square factor, take its root, and leave the rest under the radical.

Examples:

Surd	Largest perfect square factor	Simplified
$\sqrt{8}$	$4$	$\sqrt{4 \cdot 2} = 2\sqrt{2}$
$\sqrt{50}$	$25$	$5\sqrt{2}$
$\sqrt{72}$	$36$	$6\sqrt{2}$
$\sqrt{18}$	$9$	$3\sqrt{2}$
$\sqrt{45}$	$9$	$3\sqrt{5}$
$\sqrt{75}$	$25$	$5\sqrt{3}$
$\sqrt{200}$	$100$	$10\sqrt{2}$

Worked qualitative. Why pull out the LARGEST perfect square?

$\sqrt{72} = \sqrt{4 \cdot 18} = 2\sqrt{18}$ .
But $\sqrt{18}$ still simplifies: $2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}$ .
Same answer, more steps. Largest factor → done in one step.

Edexcel tip. ALWAYS check: can the radical be simplified further? If $\sqrt{18}$ appears in your answer, you missed a factor. Pull out the $9$ .

Find LARGEST perfect square.
$\sqrt{ab} = \sqrt{a}\sqrt{b}$ .
Memorise common ones: $\sqrt{8}, \sqrt{18}, \sqrt{32}, \sqrt{50}, \sqrt{72}$ .
Check radical can't simplify further.

Add, subtract, multiply

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Same radical: combine. Different: leave alone. FOIL works on surd brackets.

Addition and subtraction.

Surds with the SAME radical part combine like terms:

$5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$ .
$7\sqrt{3} - 2\sqrt{3} = 5\sqrt{3}$ .

DIFFERENT radicals can't be combined:

$\sqrt{2} + \sqrt{3}$ stays as is.

Multiplication.

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ .

Examples:

$\sqrt{2} \times \sqrt{3} = \sqrt{6}$ .
$\sqrt{8} \times \sqrt{2} = \sqrt{16} = 4$ .
$3\sqrt{2} \times 2\sqrt{5} = 6\sqrt{10}$ .

Squaring a surd.

$(\sqrt{a})^{2} = a$ .

$(\sqrt{7})^{2} = 7$ .
$(2\sqrt{3})^{2} = 4 \times 3 = 12$ .

Expanding brackets uses FOIL.

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

F: $\sqrt{3} \times \sqrt{3} = 3$ .
O: $\sqrt{3} \times (-5) = -5\sqrt{3}$ .
I: $2 \times \sqrt{3} = 2\sqrt{3}$ .
L: $2 \times (-5) = -10$ .
Combine: $3 - 5\sqrt{3} + 2\sqrt{3} - 10 = -7 - 3\sqrt{3}$ .

Worked qualitative. Compute $(\sqrt{5} + 1)^{2}$ .

FOIL or square-of-a-binomial: $(\sqrt{5})^{2} + 2 \times \sqrt{5} \times 1 + 1^{2}$ .
$= 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$ .

Edexcel tip. Show every FOIL step. Mark schemes credit method explicitly. After multiplying, simplify any surds and combine like terms.

Same radical: combine.
Different: leave alone.
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ .
$(\sqrt{a})^{2} = a$ .
FOIL for brackets.

Rationalising the denominator

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Get rid of surds in denominators. Simple: multiply by the surd. Complex: use the conjugate.

Why? Edexcel prefers (and mark schemes expect) answers with rational denominators.

Simple case: denominator is $\sqrt{c}$ .

Multiply numerator AND denominator by $\sqrt{c}$ :

$\dfrac{a}{\sqrt{c}} \times \dfrac{\sqrt{c}}{\sqrt{c}} = \dfrac{a\sqrt{c}}{c}$

Example: $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$ .

Complex case: denominator is $a + b\sqrt{c}$ or $a - b\sqrt{c}$ .

Use the CONJUGATE — same expression with the OTHER sign on the surd.

The product $(a + b\sqrt{c})(a - b\sqrt{c}) = a^{2} - b^{2}c$ has no surd.

Multiply by the conjugate: the cross terms cancel and the denominator becomes rational.

Example. $\dfrac{4}{3 + \sqrt{2}}$ .

Multiply by conjugate $\dfrac{3 - \sqrt{2}}{3 - \sqrt{2}}$ :

$\dfrac{4(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} = \dfrac{12 - 4\sqrt{2}}{9 - 2} = \dfrac{12 - 4\sqrt{2}}{7}$

Why does the conjugate work? It's the difference of two squares:

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

The middle terms cancel; what's left is rational.

Worked qualitative. Rationalise $\dfrac{2}{1 - \sqrt{3}}$ .

Conjugate: $1 + \sqrt{3}$ .
$\dfrac{2(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \dfrac{2 + 2\sqrt{3}}{1 - 3} = \dfrac{2 + 2\sqrt{3}}{-2} = -1 - \sqrt{3}$ .

Edexcel tip. Show the conjugate explicitly. Multiply both numerator and denominator. Then expand and simplify. Each step earns marks.

Simple: multiply by $\sqrt{c}/\sqrt{c}$ .
Complex: multiply by conjugate.
Conjugate flips the sign on the surd.
Difference of squares kills the surd.

Quick recap

Surd = irrational root.
Simplify by largest perfect square factor.
Same radical: combine. Different: don't.
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ . $(\sqrt{a})^{2} = a$ .
Rationalise simple: × $\sqrt{c}/\sqrt{c}$ .
Rationalise complex: × conjugate.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Surd — irrational root.
$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ .
$(\sqrt{a})^{2} = a$ .
Conjugate of $a + b\sqrt{c}$ = $a - b\sqrt{c}$ .
Rationalise: multiply num + den by conjugate.

How it’s examined

Surds appear regularly Paper 1H (3-5 marks: simplify or rationalise). Sometimes in algebra/quadratic context. Examiner reports flag (1) writing $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ , (2) leaving non-simplified surds, (3) wrong sign on conjugate.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on surds, written exactly the way a tutor would explain them at the board.

1Simplify a surd
Higher• simplify, surd
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Question
Simplify $\sqrt{72}$ .
Step-by-step solution
1. Step 1
  Find the LARGEST PERFECT SQUARE FACTOR of $72$ . $72 = 36 \times 2$ , and $36$ is a perfect square.
2. Step 2
  Use $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ :
  $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
Answer
$6\sqrt{2}$
Examiner tip
ALWAYS pull out the LARGEST perfect square factor. If you do $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$ , you'd need to simplify further. Edexcel mark schemes credit the simplest form only.
2Add and subtract surds
Higher• add, subtract
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Question
Simplify $\sqrt{50} + \sqrt{18} - \sqrt{8}$ .
Step-by-step solution
1. Step 1
  Simplify each surd: $\sqrt{50} = 5\sqrt{2}$ , $\sqrt{18} = 3\sqrt{2}$ , $\sqrt{8} = 2\sqrt{2}$ .
2. Step 2
  Now all in $\sqrt{2}$ :
  $5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2} = (5 + 3 - 2)\sqrt{2} = 6\sqrt{2}$
Answer
$6\sqrt{2}$
Examiner tip
You can only add/subtract surds with the SAME radical part. Simplify first, then combine 'like terms'.
3Multiply and expand surds
Higher• Adapted from 4MA1/1H May/Jun 2024 Q12• multiply, expand
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Question
Expand and simplify $(\sqrt{3} + 2)(\sqrt{3} - 5)$ .
Step-by-step solution
1. Step 1
  FOIL — multiply each pair:
  $\sqrt{3} \times \sqrt{3} - 5\sqrt{3} + 2\sqrt{3} - 10$
2. Step 2
  $\sqrt{3} \times \sqrt{3} = 3$ . Combine like terms:
  $3 - 3\sqrt{3} - 10 = -7 - 3\sqrt{3}$
Answer
$-7 - 3\sqrt{3}$
Examiner tip
FOIL applies to surd brackets just like algebraic ones. $\sqrt{a} \times \sqrt{a} = a$ — the surd disappears.
4Rationalise a simple denominator
Higher• rationalising
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Question
Rationalise the denominator of $\dfrac{6}{\sqrt{3}}$ .
Step-by-step solution
1. Step 1
  Multiply numerator AND denominator by $\sqrt{3}$ :
  $\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3}$
2. Step 2
  Simplify: $\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$ .
Answer
$2\sqrt{3}$
Examiner tip
$\sqrt{a} \times \sqrt{a} = a$ kills the surd in the denominator. Always simplify the result.
5Rationalise using a conjugate
Higher• Adapted from 4MA1/1H Jan 2024 Q19• rationalising, conjugate
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Question
Rationalise the denominator of $\dfrac{4}{3 + \sqrt{2}}$ , giving your answer in the form $a + b\sqrt{2}$ .
Step-by-step solution
1. Step 1
  Multiply by the CONJUGATE: $3 - \sqrt{2}$ .
  $\frac{4}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{4(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})}$
2. Step 2
  Denominator: difference of two squares $= 3^{2} - (\sqrt{2})^{2} = 9 - 2 = 7$ .
3. Step 3
  Numerator: $4(3 - \sqrt{2}) = 12 - 4\sqrt{2}$ .
4. Step 4
  Combine:
  $\frac{12 - 4\sqrt{2}}{7} = \frac{12}{7} - \frac{4}{7}\sqrt{2}$
Answer
$\dfrac{12}{7} - \dfrac{4}{7}\sqrt{2}$
Examiner tip
The CONJUGATE of $a + b\sqrt{c}$ is $a - b\sqrt{c}$ . Multiplying gives $(a)^{2} - (b\sqrt{c})^{2}$ — a difference of squares with no surd.
6Recognising which numbers are surds [1 mark]
Getting started• Paper 1H/2H style — 1 mark• surds, meaning of a surd
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Question
Here are four numbers.

$\sqrt{25}$ $\qquad$ $\sqrt{30}$ $\qquad$ $\sqrt{\dfrac{49}{4}}$ $\qquad$ $2\sqrt{3}$

Write down the two numbers that are surds. [1]
Step-by-step solution
1. Step 1
  A surd is a root that cannot be written exactly as a whole number or a fraction, so it is irrational. Test each number.
  $\sqrt{25} = 5, \qquad \sqrt{\tfrac{49}{4}} = \tfrac{7}{2}$
2. Step 2
  B1 $30$ is not a square number, so $\sqrt{30}$ is irrational. $2\sqrt{3}$ is a whole number times an irrational root, so it is also a surd.
Answer
$\sqrt{30}$ and $2\sqrt{3}$
Examiner tip
Examiner tip: A root sign alone does not make a surd. Ask whether the root can be worked out exactly. $\sqrt{\tfrac{49}{4}} = 3.5$ , so it is rational.

Common mistake: Leaving out $2\sqrt{3}$ because it has a whole number in front. Multiplying a surd by a whole number still gives a surd. Both answers are needed for the mark.
7Rationalising a denominator with a coefficient [2 marks]
Getting started• Paper 1H/2H style — 2 marks• rationalising, surds
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Question
Write $\dfrac{9}{2\sqrt{3}}$ in the form $\dfrac{a\sqrt{3}}{b}$ , where $a$ and $b$ are integers. Give your answer in its simplest form. Show your working clearly. [2]
Step-by-step solution
1. Step 1
  M1 Multiply top and bottom by $\sqrt{3}$ only. The $2$ is already rational, so it does not need to be removed.
  $\frac{9}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{9\sqrt{3}}{2 \times 3} = \frac{9\sqrt{3}}{6}$
2. Step 2
  A1 Cancel the common factor of $3$ .
  $\frac{9\sqrt{3}}{6} = \frac{3\sqrt{3}}{2}$
Answer
$\dfrac{3\sqrt{3}}{2}$
Examiner tip
Examiner tip: In the denominator, $2\sqrt{3} \times \sqrt{3} = 2 \times 3 = 6$ . Multiply the whole number as well.

Common mistake: Writing the denominator as $2 \times \sqrt{3} \times \sqrt{3} = 2\sqrt{3}$ , or stopping at $\dfrac{9\sqrt{3}}{6}$ . The first is wrong and the second is not in its simplest form.
8Expanding brackets, then collecting another surd [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• expanding brackets, surds
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Question
Express $(2\sqrt{5} - 3)(\sqrt{5} + 4) - \sqrt{45}$ in the form $a + b\sqrt{5}$ , where $a$ and $b$ are integers. Show your working clearly. [3]
Step-by-step solution
1. Step 1
  M1 Expand the brackets (four terms, at least three correct). Note $2\sqrt{5} \times \sqrt{5} = 2 \times 5 = 10$ .
  $(2\sqrt{5} - 3)(\sqrt{5} + 4) = 10 + 8\sqrt{5} - 3\sqrt{5} - 12 = -2 + 5\sqrt{5}$
2. Step 2
  B1 Simplify $\sqrt{45}$ so it can be collected with the other $\sqrt{5}$ terms.
  $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$
3. Step 3
  A1 Collect the like surds.
  $-2 + 5\sqrt{5} - 3\sqrt{5} = -2 + 2\sqrt{5}$
Answer
$-2 + 2\sqrt{5}$ , so $a = -2$ , $b = 2$
Examiner tip
Examiner tip: The B1 for $\sqrt{45} = 3\sqrt{5}$ is independent of the expansion, so write it down even if the brackets go wrong.

Common mistake: Writing $2\sqrt{5} \times \sqrt{5} = 2\sqrt{25}$ and leaving it there, or getting $+12$ instead of $-12$ from $-3 \times 4$ . A decimal from the calculator ( $2.472\ldots$ ) scores 0 because working is required.
9Rationalising to reach the form a + b√7 [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• rationalising, conjugate, surds
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Question
Express $\dfrac{1 + \sqrt{7}}{3 - \sqrt{7}}$ in the form $a + b\sqrt{7}$ , where $a$ and $b$ are integers. Show your working clearly. [4]
Step-by-step solution
1. Step 1
  M1 Multiply top and bottom by the conjugate $3 + \sqrt{7}$ .
  $\frac{1 + \sqrt{7}}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}}$
2. Step 2
  M1 Denominator: difference of two squares.
  $(3 - \sqrt{7})(3 + \sqrt{7}) = 9 - 7 = 2$
3. Step 3
  M1 Numerator: four terms, then collect.
  $(1 + \sqrt{7})(3 + \sqrt{7}) = 3 + \sqrt{7} + 3\sqrt{7} + 7 = 10 + 4\sqrt{7}$
4. Step 4
  A1 Divide both terms by $2$ .
  $\frac{10 + 4\sqrt{7}}{2} = 5 + 2\sqrt{7}$
Answer
$5 + 2\sqrt{7}$
Examiner tip
Examiner tip: Check with the calculator at the end: both the question and $5 + 2\sqrt{7}$ give $10.29\ldots$ . The calculator confirms the answer, and the written working earns the marks.

Common mistake: Stopping at $\dfrac{10 + 4\sqrt{7}}{2}$ . It is the right value but not in the form $a + b\sqrt{7}$ , so the A1 is lost. Examiners often report this: candidates stop before reaching the required form.
10Area and diagonal of a rectangle with surd sides [5 marks]
Stretch• Paper 1H/2H style — 5 marks• surds in context, pythagoras, surds
▼
Question
A rectangle has length $(5 + \sqrt{3})$ cm and width $(5 - \sqrt{3})$ cm.

(a) Work out the area of the rectangle. Show your working clearly. [2]

(b) Find the exact length of a diagonal of the rectangle. Give your answer as a surd in its simplest form. [3]
Step-by-step solution
1. Step 1
  M1 (a) Multiply length by width. This is a difference of two squares.
  $(5 + \sqrt{3})(5 - \sqrt{3}) = 25 - 5\sqrt{3} + 5\sqrt{3} - 3$
2. Step 2
  A1 The surd terms cancel.
  $25 - 3 = 22 \text{ cm}^{2}$
3. Step 3
  M1 (b) The diagonal is the hypotenuse of a right-angled triangle whose shorter sides are the length and the width. Square each side in full.
  $(5 + \sqrt{3})^{2} = 28 + 10\sqrt{3}, \qquad (5 - \sqrt{3})^{2} = 28 - 10\sqrt{3}$
4. Step 4
  M1 Add the squares. The surd terms cancel again.
  $d^{2} = 28 + 10\sqrt{3} + 28 - 10\sqrt{3} = 56$
5. Step 5
  A1 Square root, then simplify using $56 = 4 \times 14$ .
  $d = \sqrt{56} = 2\sqrt{14}$
Answer
(a) $22$ cm² (b) $2\sqrt{14}$ cm
Examiner tip
Examiner tip: Pythagoras needs the squares of the sides, so square each bracket in full, middle term included, before adding.

Common mistake: Writing $(5 + \sqrt{3})^{2} = 25 + 3$ . This is the same error as $(\sqrt{a} + \sqrt{b})^{2} = a + b$ . Here it happens to give the same total, $56$ , because the middle terms cancel. That is luck, and the M1 for squaring is not awarded for a wrong expansion. Also, $7.48$ cm is not exact and loses the A1.

Question 1
Paper 1H/2H style — 1 mark1 mark
Write down the value of $(2\sqrt{7})^{2}$ . [1]
Model answer
$(2\sqrt{7})^{2} = 2^{2} \times (\sqrt{7})^{2} = 4 \times 7 = 28$

Answer: $28$
Why this scores
Mark scheme:
- B1 for $28$ cao
Why this scores full marks: The square applies to both factors, the $2$ and the $\sqrt{7}$ . Answers of $14$ (only the root squared) and $4\sqrt{7}$ (only the $2$ squared) are the two usual wrong answers.
Question 2
Paper 1H/2H style — 2 marks2 marks
$\sqrt{12} \times \sqrt{n} = 30$ , where $n$ is an integer.

Find the value of $n$ . [2]
Model answer
$\sqrt{12} \times \sqrt{n} = \sqrt{12n}$

$\sqrt{12n} = 30 \Rightarrow 12n = 900$

$n = 75$
Why this scores
Mark scheme:
- M1 for $12n = 900$ oe, or $\sqrt{n} = \dfrac{30}{2\sqrt{3}}$ oe
- A1 for $75$ cao
Why this scores full marks: Combining into one root and then squaring both sides turns the question into a one-step equation. Check: $\sqrt{12 \times 75} = \sqrt{900} = 30$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
Show that $(\sqrt{2} + \sqrt{8})^{2} = 18$ . [2]
Model answer
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

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

$(3\sqrt{2})^{2} = 9 \times 2 = 18$
Why this scores
Mark scheme:
- M1 for $\sqrt{8} = 2\sqrt{2}$ , or for the expansion $2 + \sqrt{16} + \sqrt{16} + 8$ oe
- A1 for a complete, correct argument leading to $18$
Why this scores full marks: The candidate simplifies first, so the square is easy. The trap is $(\sqrt{2} + \sqrt{8})^{2} = 2 + 8 = 10$ , which drops the middle term, a mistake examiners report every year. Expanding four terms is equally valid: $2 + 4 + 4 + 8 = 18$ .
Question 4
Paper 1H/2H style — 2 marks2 marks
Rationalise the denominator of $\dfrac{\sqrt{10} + 5}{\sqrt{5}}$ . Give your answer in the form $\sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are integers. Show your working clearly. [2]
Model answer
$\dfrac{\sqrt{10} + 5}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{\sqrt{50} + 5\sqrt{5}}{5}$

$= \dfrac{5\sqrt{2} + 5\sqrt{5}}{5} = \sqrt{2} + \sqrt{5}$
Why this scores
Mark scheme:
- M1 for multiplying numerator and denominator by $\sqrt{5}$ , or for $\dfrac{\sqrt{10}}{\sqrt{5}} + \dfrac{5}{\sqrt{5}}$
- A1 for $\sqrt{2} + \sqrt{5}$ cao
Why this scores full marks: The whole numerator is multiplied by $\sqrt{5}$ , not just the first term, and $\sqrt{50}$ is simplified to $5\sqrt{2}$ so the $5$ cancels. Stopping at $\dfrac{\sqrt{50} + 5\sqrt{5}}{5}$ is not in the required form.
Question 5
Paper 1H/2H style — 3 marks3 marks
Expand and simplify $(3\sqrt{2} + \sqrt{6})(\sqrt{2} - \sqrt{6})$ . Give your answer in the form $k\sqrt{3}$ , where $k$ is an integer. Show your working clearly. [3]
Model answer
$(3\sqrt{2} + \sqrt{6})(\sqrt{2} - \sqrt{6}) = 3 \times 2 - 3\sqrt{12} + \sqrt{12} - 6$

$= 6 - 6 - 2\sqrt{12} = -2\sqrt{12}$

$\sqrt{12} = 2\sqrt{3}$ , so $-2\sqrt{12} = -4\sqrt{3}$

Answer: $-4\sqrt{3}$
Why this scores
Mark scheme:
- M1 for $6 - 3\sqrt{12} + \sqrt{12} - 6$ (at least 3 of the 4 terms correct)
- M1 for $-2\sqrt{12}$ or $\sqrt{12} = 2\sqrt{3}$
- A1 for $-4\sqrt{3}$ cao
Why this scores full marks: Each product is worked out correctly: $3\sqrt{2} \times \sqrt{2} = 6$ , $\sqrt{2} \times \sqrt{6} = \sqrt{12}$ and $\sqrt{6} \times \sqrt{6} = 6$ . The whole numbers cancel, which is easy to miss, and the answer is taken all the way to the form $k\sqrt{3}$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
Rationalise the denominator of $\dfrac{10}{\sqrt{7} - \sqrt{2}}$ . Simplify your answer fully. Show your working clearly. [3]
Model answer
$\dfrac{10}{\sqrt{7} - \sqrt{2}} \times \dfrac{\sqrt{7} + \sqrt{2}}{\sqrt{7} + \sqrt{2}} = \dfrac{10(\sqrt{7} + \sqrt{2})}{7 - 2}$

$= \dfrac{10(\sqrt{7} + \sqrt{2})}{5} = 2(\sqrt{7} + \sqrt{2})$

Answer: $2\sqrt{7} + 2\sqrt{2}$
Why this scores
Mark scheme:
- M1 for multiplying numerator and denominator by $\sqrt{7} + \sqrt{2}$
- M1 for a denominator of $7 - 2$ or $5$
- A1 for $2\sqrt{7} + 2\sqrt{2}$ oe, e.g. $2(\sqrt{7} + \sqrt{2})$
Why this scores full marks: The conjugate changes only the sign in the middle. The denominator is then $(\sqrt{7})^{2} - (\sqrt{2})^{2} = 5$ , not $7 + 2$ . The final $\div 5$ is carried out, so the answer is fully simplified.
Question 7
Paper 1H/2H style — 3 marks3 marks
A triangle has sides of length $\sqrt{18}$ cm, $\sqrt{32}$ cm and $\sqrt{50}$ cm.

(a) Show that the perimeter of the triangle is $12\sqrt{2}$ cm. [2]

(b) Explain why the triangle is right-angled. [1]
Model answer
(a) $\sqrt{18} = 3\sqrt{2}$ , $\sqrt{32} = 4\sqrt{2}$ , $\sqrt{50} = 5\sqrt{2}$

$3\sqrt{2} + 4\sqrt{2} + 5\sqrt{2} = 12\sqrt{2}$ cm

(b) $(\sqrt{18})^{2} + (\sqrt{32})^{2} = 18 + 32 = 50 = (\sqrt{50})^{2}$ , so the sides satisfy Pythagoras' theorem and the triangle is right-angled.
Why this scores
Mark scheme:
- M1 for at least two of $3\sqrt{2}$ , $4\sqrt{2}$ , $5\sqrt{2}$
- A1 for a complete, correct argument leading to $12\sqrt{2}$
- B1 for $18 + 32 = 50$ with a reference to Pythagoras oe
Why this scores full marks: In (a) every side is written as a multiple of $\sqrt{2}$ before adding. $\sqrt{18 + 32 + 50} = 10$ is the usual wrong answer. In (b) the numbers are shown and the conclusion is stated. A bare "it satisfies Pythagoras" with no numbers does not get the B1.
Question 8
Paper 1H/2H style — 4 marks4 marks
$p = 2 + \sqrt{3}$

Find the value of $p^{2} + \dfrac{1}{p^{2}}$ . Show your working clearly. [4]
Model answer
$\dfrac{1}{p} = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$

$p^{2} = (2 + \sqrt{3})^{2} = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$

$\dfrac{1}{p^{2}} = (2 - \sqrt{3})^{2} = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$

$p^{2} + \dfrac{1}{p^{2}} = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14$
Why this scores
Mark scheme:
- M1 for rationalising $\dfrac{1}{2 + \sqrt{3}}$ using $2 - \sqrt{3}$ , or for $\dfrac{1}{7 + 4\sqrt{3}} \times \dfrac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}}$
- A1 for $\dfrac{1}{p} = 2 - \sqrt{3}$ or $\dfrac{1}{p^{2}} = 7 - 4\sqrt{3}$
- M1 for expanding $(2 + \sqrt{3})^{2}$ correctly, giving $7 + 4\sqrt{3}$
- A1 for $14$ cao
Why this scores full marks: The candidate notices that the denominator $4 - 3$ is $1$ , so the reciprocal is just the conjugate. After that, two short expansions and the surds cancel. A calculator value of $14$ with no working scores 0.
Question 9
Paper 1H/2H style — 4 marks4 marks
Find the exact value of $\dfrac{\sqrt{5} - 1}{\sqrt{5} + 1} + \dfrac{\sqrt{5} + 1}{\sqrt{5} - 1}$ . Show your working clearly. [4]
Model answer
Common denominator: $(\sqrt{5} + 1)(\sqrt{5} - 1) = 5 - 1 = 4$

$\dfrac{(\sqrt{5} - 1)^{2} + (\sqrt{5} + 1)^{2}}{4}$

$(\sqrt{5} - 1)^{2} = 6 - 2\sqrt{5}$ and $(\sqrt{5} + 1)^{2} = 6 + 2\sqrt{5}$

$\dfrac{6 - 2\sqrt{5} + 6 + 2\sqrt{5}}{4} = \dfrac{12}{4} = 3$
Why this scores
Mark scheme:
- M1 for a common denominator of $(\sqrt{5} + 1)(\sqrt{5} - 1)$ , or for rationalising either fraction with the correct conjugate
- M1 for a denominator of $4$
- M1 for $6 - 2\sqrt{5}$ and $6 + 2\sqrt{5}$ oe
- A1 for $3$ cao
Why this scores full marks: The candidate sees that each denominator is the other one's conjugate, so a single common denominator rationalises both at once. Each square is expanded with its middle term, and the answer is exact.
Question 10
Paper 1H/2H style — 5 marks5 marks
In a right-angled triangle, the hypotenuse has length $(4 + \sqrt{2})$ cm and one of the other sides has length $(2 + 2\sqrt{2})$ cm.

(a) Find the exact length of the third side. Show your working clearly. [3]

(b) Find the area of the triangle. Give your answer in the form $(\sqrt{a} + b\sqrt{3})$ cm², where $a$ and $b$ are integers. Show your working clearly. [2]
Model answer
(a) $(4 + \sqrt{2})^{2} = 16 + 8\sqrt{2} + 2 = 18 + 8\sqrt{2}$

$(2 + 2\sqrt{2})^{2} = 4 + 8\sqrt{2} + 8 = 12 + 8\sqrt{2}$

$x^{2} = (18 + 8\sqrt{2}) - (12 + 8\sqrt{2}) = 6$

$x = \sqrt{6}$ cm

(b) Area $= \tfrac{1}{2}(2 + 2\sqrt{2})\sqrt{6} = (1 + \sqrt{2})\sqrt{6} = \sqrt{6} + \sqrt{12}$

$= \sqrt{6} + 2\sqrt{3}$ cm²
Why this scores
Mark scheme:
- M1 for expanding either square correctly, $18 + 8\sqrt{2}$ or $12 + 8\sqrt{2}$
- M1 for $(18 + 8\sqrt{2}) - (12 + 8\sqrt{2})$ oe
- A1 for $\sqrt{6}$ cao
- M1 for $\tfrac{1}{2} \times (2 + 2\sqrt{2}) \times$ their $\sqrt{6}$
- A1 for $\sqrt{6} + 2\sqrt{3}$ cao
Why this scores full marks: The hypotenuse is identified correctly, so the squares are subtracted, not added. Each bracket is squared with its middle term, and the $8\sqrt{2}$ terms cancel exactly. In (b) the two sides at the right angle are the base and the height, and $\sqrt{12}$ is simplified to reach the required form.

4MA1 Surds

At a glance

What you’ll learn

Simplify surds

Add, subtract, multiply

Rationalising the denominator

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Simplify a surd

2Add and subtract surds

3Multiply and expand surds

4Rationalise a simple denominator

5Rationalise using a conjugate

6Recognising which numbers are surds [1 mark]

7Rationalising a denominator with a coefficient [2 marks]

8Expanding brackets, then collecting another surd [3 marks]

9Rationalising to reach the form a + b√7 [4 marks]

10Area and diagonal of a rectangle with surd sides [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10