Number

Surds

14 past-paper questions, easiest first. Try each one, then reveal the answer.

Higher only

Revision guide by The Full Marks

Surds: what you need to know

A surd is a square root that cannot be simplified to a whole number or fraction, such as $\sqrt{2}$ . Surds let you give exact answers, and Higher papers test simplifying them and rationalising denominators, usually on the non-calculator paper.

Key facts

$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ .
$\sqrt{a} \times \sqrt{a} = a$ , so $3\sqrt{2} \times 3\sqrt{2} = 18$ .
You can only add like surds: $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ , but $\sqrt{2} + \sqrt{3}$ stays as it is.
$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ , which has no surd.

Simplifying a surd

Find the largest square number that is a factor of the number under the root.
Split the root: $\sqrt{48} = \sqrt{16} \times \sqrt{3}$ .
Square root the square number: $4\sqrt{3}$ .

Expanding brackets with surds

Multiply every term in the first bracket by every term in the second, as with $(x + 3)(x + 5)$ .
Use $\sqrt{a} \times \sqrt{a} = a$ and $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ .
Collect the whole numbers and the surd terms separately: $(3 + \sqrt{5})^2 = 9 + 3\sqrt{5} + 3\sqrt{5} + 5 = 14 + 6\sqrt{5}$ .

Rationalising the denominator

If the bottom is $\sqrt{b}$ , multiply top and bottom by $\sqrt{b}$ .
If the bottom is $a - \sqrt{b}$ , multiply top and bottom by $a + \sqrt{b}$ (swap the sign).
Expand, simplify and cancel any common factors.

Worked examples

Example 1 Higher · grade 6

Simplify $\sqrt{75} + \sqrt{12}$

Show the solution

$\sqrt{75} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$
$\sqrt{12} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$
$5\sqrt{3} + 2\sqrt{3} = 7\sqrt{3}$

Example 2 Higher · grade 8

Write $\frac{6}{3 - \sqrt{3}}$ in the form $a + \sqrt{b}$ , where $a$ and $b$ are integers.

Show the solution

Multiply top and bottom by $3 + \sqrt{3}$ :
Top: $6(3 + \sqrt{3}) = 18 + 6\sqrt{3}$
Bottom: $(3 - \sqrt{3})(3 + \sqrt{3}) = 9 - 3 = 6$
$\frac{18 + 6\sqrt{3}}{6} = 3 + \sqrt{3}$
So $a = 3$ and $b = 3$

Common mistakes

Adding under the root: $\sqrt{8} + \sqrt{2}$ is not $\sqrt{10}$ . Simplify first: $2\sqrt{2} + \sqrt{2} = 3\sqrt{2}$ .
Multiplying by the same bracket instead of the one with the opposite sign when rationalising.
Forgetting to square the whole number: $(2\sqrt{3})^2 = 4 \times 3 = 12$ , not 6.
Squaring a bracket term by term: $(3 + \sqrt{5})^2$ is not $9 + 5$ . Write it as $(3 + \sqrt{5})(3 + \sqrt{5})$ and include the middle terms.

Exam tips

These are usually non-calculator questions, and you must show the surd steps to get the marks.
Check the form the question asks for, e.g. $a\sqrt{b}$ , and state the values of $a$ and $b$ .
Simplify any surd such as $\sqrt{180} = 6\sqrt{5}$ before expanding or rationalising; it keeps the numbers small. In a "show that" question, write every line of working down to the exact form given.

Questions people ask

What is a surd?

A surd is a root that cannot be written as a whole number or fraction, like $\sqrt{5}$ . Leaving it as a surd keeps the answer exact.

How do you rationalise the denominator?

Multiply the top and bottom of the fraction by something that removes the root from the bottom: $\sqrt{b}$ for a single surd, or the bracket with the opposite sign for $a \pm \sqrt{b}$ .

Are surds on the Foundation paper?

No. Simplifying surds and rationalising denominators are Higher tier only.

Exam questions: Surds (14, easiest first)

Surds3 marksGrade 7No calculatorHigher

June 2024 · Paper 1 Higher · Q17

Rationalise the denominator of 1 over root 7, and simplify a difference of surds.

Show answer

(a)	$\frac{\sqrt{7}}{7}$	1 mark
(b)	$3\sqrt{5}$	2 marks

Worked solution Checked against the mark scheme

(a) Multiply top and bottom by $\sqrt{7}$ : $\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

(b) $\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$

$4\sqrt{5} - \sqrt{5} = 3\sqrt{5}$

Watch out: Writing $\sqrt{80} - \sqrt{5} = \sqrt{75}$ .

Surds2 marksGrade 7CalculatorHigher

June 2024 · Paper 2 Higher · Q15

Rationalise the denominator of 1 over (root a plus 1), where a is prime.

Show answer

Answer	$\dfrac{\sqrt a - 1}{a - 1}$	2 marks

Worked solution Checked against the mark scheme

Multiply top and bottom by $\sqrt a - 1$ :

$\dfrac{1}{\sqrt a + 1} \times \dfrac{\sqrt a - 1}{\sqrt a - 1} = \dfrac{\sqrt a - 1}{a - 1}$

Watch out: Multiplying by $\sqrt a + 1$ or by $\sqrt a$ , which does not remove the surd.

Surds4 marksGrade 7No calculatorHigher

June 2023 · Paper 1 Higher · Q16

$Given a fraction of a sphere's surface area, find its diameter as a simplified surd.$

Show answer

Answer	$10\sqrt{2}\text{ cm}$	4 marks

Worked solution Checked against the mark scheme

Whole surface area $= 75\pi \div 3 \times 8 = 200\pi$

$4\pi r^2 = 200\pi$ , so $r^2 = 50$ and $r = \sqrt{50} = 5\sqrt{2}$

Diameter $= 2 \times 5\sqrt{2} = 10\sqrt{2}$ cm

Watch out: Giving the radius $5\sqrt{2}$ instead of the diameter.

Surds5 marksGrade 7No calculatorHigher

June 2019 · Paper 1 Higher · Q18

$Simplify a sum of surds, then write a power of a surd fraction with a rational denominator.$

Show answer

(a)	$3\sqrt{3}$	2 marks
(b)	$\frac{\sqrt{3}}{81}$	3 marks

Worked solution Checked against the mark scheme

(a) $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ , so $\sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$

(b) $(\sqrt{3})^7 = (\sqrt{3})^6 \times \sqrt{3} = 27\sqrt{3}$

$\frac{1}{27\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{81}$

Watch out: In (a), adding under the root to get $\sqrt{15}$ .

Surds3 marksGrade 7No calculatorHigher

June 2018 · Paper 1 Higher · Q13

Simplify a surd expression to the form a√10 and find a.

Show answer

Answer	$a = 5$	3 marks

Worked solution Checked against the mark scheme

$\sqrt{8} = 2\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$ , so $\sqrt{8} + \sqrt{18} = 5\sqrt{2}$

$\sqrt{5} \times 5\sqrt{2} = 5\sqrt{10}$

So $a = 5$

Watch out: Adding the numbers under the roots: $\sqrt{8} + \sqrt{18} = \sqrt{26}$ .

Surds2 marksGrade 7CalculatorHigher

June 2018 · Paper 2 Higher · Q20

Spot the errors in two worked attempts at rationalising a denominator.

Show answer

(a)	$\sqrt{3} \times -\sqrt{3}$ is $-3$ , not $+3$ , so the denominator should be 1	1 mark
(b)	$\sqrt{12} = 2\sqrt{3}$ , not $3\sqrt{2}$	1 mark

Worked solution Checked against the mark scheme

(a) $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$ . Martin wrote $+3$ for $\sqrt{3} \times -\sqrt{3}$ . The answer should be $28 - 14\sqrt{3}$ .

(b) $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ . Sian wrote $3\sqrt{2}$ . The answer should be $\frac{10\sqrt{3}}{12} = \frac{5\sqrt{3}}{6}$ .

Watch out: Describing the general method instead of pointing to the exact wrong step.

Surds6 marksGrade 8No calculatorHigher

November 2024 · Paper 1 Higher · Q16

$Rationalise a simple surd denominator, then rewrite a surd fraction with a two-term denominator in a given form.$

Show answer

(a)	$3\sqrt{5}$	2 marks
(b)	$\frac{32 - 9\sqrt{3}}{11}$	4 marks

Worked solution Checked against the mark scheme

(a) $\frac{15}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}$

(b) $\sqrt{75} = 5\sqrt{3}$ . Multiply top and bottom by $1 - 2\sqrt{3}$ :

Top: $(5\sqrt{3} - 2)(1 - 2\sqrt{3}) = 5\sqrt{3} - 30 - 2 + 4\sqrt{3} = 9\sqrt{3} - 32$

Bottom: $1 - 12 = -11$

$\frac{9\sqrt{3} - 32}{-11} = \frac{32 - 9\sqrt{3}}{11}$

Watch out: Getting $2\sqrt{3} \times 2\sqrt{3} = 6$ instead of 12.

Surds4 marksGrade 8No calculatorHigher

November 2021 · Paper 1 Higher · Q19

$Simplify a surd and rationalise a denominator to write a surd fraction in a stated form.$

Show answer

Answer	$\frac{17 + \sqrt{3}}{11}$	4 marks

Worked solution Checked against the mark scheme

$\sqrt{12} = 2\sqrt{3}$

Multiply top and bottom by $5 - \sqrt{3}$ :

Top: $(8 + 2\sqrt{3})(5 - \sqrt{3}) = 40 - 8\sqrt{3} + 10\sqrt{3} - 6 = 34 + 2\sqrt{3}$

Bottom: $25 - 3 = 22$

$\frac{34 + 2\sqrt{3}}{22} = \frac{17 + \sqrt{3}}{11}$

Watch out: Getting $2\sqrt{3} \times \sqrt{3} = 2\sqrt{3}$ instead of 6.

Surds4 marksGrade 8No calculatorHigher

November 2020 · Paper 1 Higher · Q20

$Simplify a surd fraction by rationalising the denominator to reach a given form.$

Show answer

Answer	$1 + \frac{\sqrt{5}}{5}$ (a = 1, b = 5)	4 marks

Worked solution Checked against the mark scheme

$\sqrt{180} = 6\sqrt{5}$ , so the top is $6\sqrt{5} - 2\sqrt{5} = 4\sqrt{5}$

Multiply top and bottom by $5\sqrt{5} + 5$ :

Top: $4\sqrt{5}(5\sqrt{5} + 5) = 100 + 20\sqrt{5}$

Bottom: $(5\sqrt{5})^2 - 5^2 = 125 - 25 = 100$

So the fraction is $\frac{100 + 20\sqrt{5}}{100} = 1 + \frac{\sqrt{5}}{5}$

Watch out: Forgetting to simplify $\sqrt{180}$ first, which makes the expansion much harder.

Surds5 marksGrade 8No calculatorHigher

November 2019 · Paper 1 Higher · Q16

$Rationalise a simple surd denominator, then show a fraction with a two-term surd denominator can be written in a given form.$

Show answer

(a)	$2\sqrt{11}$	2 marks
(b)	$\frac{6+\sqrt{3}}{11}$ , so $a = 6$ , $b = 11$	3 marks

Worked solution Checked against the mark scheme

(a) $\frac{22}{\sqrt{11}} \times \frac{\sqrt{11}}{\sqrt{11}} = \frac{22\sqrt{11}}{11} = 2\sqrt{11}$

(b) Multiply by $\frac{2\sqrt{3}+1}{2\sqrt{3}+1}$ :

Top: $\sqrt{3}(2\sqrt{3}+1) = 6 + \sqrt{3}$

Bottom: $(2\sqrt{3})^2 - 1^2 = 12 - 1 = 11$

So it equals $\frac{6+\sqrt{3}}{11}$ , with $a = 6$ and $b = 11$ .

Watch out: In (b), multiplying by 2√3 − 1 instead of 2√3 + 1.

Surds3 marksGrade 8No calculatorHigher

November 2018 · Paper 1 Higher · Q20

Expand and simplify a surd expression and rationalise the denominator to reach the form a(b + √2).

Show answer

Answer	$16(1 + \sqrt{2})$ , so $a = 16$ , $b = 1$	3 marks

Worked solution Checked against the mark scheme

$\sqrt{18} + \sqrt{2} = 3\sqrt{2} + \sqrt{2} = 4\sqrt{2}$ , so the numerator is $(4\sqrt{2})^2 = 32$

$\sqrt{8} - 2 = 2\sqrt{2} - 2$

$\dfrac{32}{2\sqrt{2} - 2} \times \dfrac{2\sqrt{2} + 2}{2\sqrt{2} + 2} = \dfrac{32(2\sqrt{2} + 2)}{8 - 4} = 8(2\sqrt{2} + 2)$

$= 16\sqrt{2} + 16 = 16(1 + \sqrt{2})$

Watch out: Squaring the numerator as $18 + 2 = 20$ and forgetting the middle terms.

Surds3 marksGrade 8No calculatorHigher

Specimen · Paper 1 Higher · Q19

$Expand a product of surd brackets and simplify the resulting fraction by rationalising the denominator.$

Show answer

Answer	$\sqrt{31}$	3 marks

Worked solution Checked against the mark scheme

Expand the numerator: $(6 - \sqrt{5})(6 + \sqrt{5}) = 36 + 6\sqrt{5} - 6\sqrt{5} - 5 = 31$

So the expression is $\frac{31}{\sqrt{31}}$ .

Multiply top and bottom by $\sqrt{31}$ : $\frac{31\sqrt{31}}{31} = \sqrt{31}$

Watch out: Expanding to $36 - 5\sqrt{5}$ or $36 + 5$ .

Surds4 marksGrade 9No calculatorHigher

June 2023 · Paper 1 Higher · Q23

$Rationalise and combine two surd fractions into a single fraction of a given form.$

Show answer

Answer	$\frac{10\sqrt{3} + 27}{39}$	4 marks

Worked solution Checked against the mark scheme

$\frac{3\sqrt{3}}{4 - \sqrt{3}} \times \frac{4 + \sqrt{3}}{4 + \sqrt{3}} = \frac{12\sqrt{3} + 9}{13}$

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

Common denominator 39: $\frac{36\sqrt{3} + 27}{39} - \frac{26\sqrt{3}}{39} = \frac{10\sqrt{3} + 27}{39}$

Watch out: Multiplying by $4 - \sqrt{3}$ instead of the conjugate $4 + \sqrt{3}$ .

Surds4 marksGrade 9No calculatorHigher

November 2023 · Paper 1 Higher · Q22

Find the common ratio of a geometric sequence whose terms involve surds, rationalising the denominator.

Show answer

Answer	$3 + \sqrt{2}$	4 marks

Worked solution Checked against the mark scheme

$r = \frac{13 + 9\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}$

Numerator: $39 - 26\sqrt{2} + 27\sqrt{2} - 36 = 3 + \sqrt{2}$

Denominator: $9 - 8 = 1$

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

Watch out: Dividing the 2nd term by the 3rd (upside down), or errors with $\sqrt{2} \times \sqrt{2}$ .

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2024 · Paper 1 Foundation · Q20: Subtract mixed numbers, then show a division of mixed numbers gives a stated result.
June 2024 · Paper 1 Foundation · Q21: Subtract mixed numbers and spot the error in a worked multiplication of fractions.
June 2024 · Paper 3 Higher · Q21: Compare the perimeters of inscribed and circumscribed regular hexagons with a circle's circumference to show bounds for pi.
November 2023 · Paper 1 Foundation · Q17: Divide one proper fraction by another and write the result as a mixed number.
November 2023 · Paper 1 Higher · Q20: Use the cosine rule with a surd side length to find an angle in a triangle without a calculator.
June 2023 · Paper 1 Higher · Q2: Subtract one mixed number from another and give a mixed number.
June 2023 · Paper 1 Foundation · Q21: Subtract one mixed number from another and give a mixed number.
June 2023 · Paper 2 Higher · Q24: Find the exact area of a star-shaped region enclosed by eight regular octagons, in surd form.
November 2022 · Paper 1 Higher · Q19: Solve an equation involving two algebraic fractions, giving the solutions in surd form.
November 2022 · Paper 1 Foundation · Q20: Add two mixed numbers, then show a mixed number divided by a whole number gives a stated fraction.
November 2022 · Paper 1 Higher · Q23: Find the next term of a geometric sequence with surds, then find the first term of another from its 4th and 6th terms.
November 2022 · Paper 1 Higher · Q24: Find the ratio of radius to height when a sphere and cone have equal volumes, then when a sphere's surface area equals a cone's total surface area.
June 2022 · Paper 1 Foundation · Q12: Add two fractions with different denominators and multiply two fractions, simplifying fully.
June 2022 · Paper 1 Higher · Q18: Evaluate an expression with a negative fractional power of a mixed number, a mixed-number product and a negative power of 2.
June 2022 · Paper 1 Higher · Q21: Find the exact area of the two shaded regions formed inside the middle of three overlapping equal circles.
June 2022 · Paper 2 Foundation · Q2: Find the fraction that makes one minus three tenths correct.
November 2021 · Paper 1 Foundation · Q22: Subtract two mixed numbers and give the result as a mixed number.
November 2021 · Paper 3 Foundation · Q14: Find the remaining part of a cost after a given fraction has been spent.
November 2019 · Paper 1 Foundation · Q22: Multiply two mixed numbers and give the answer as a mixed number.
June 2018 · Paper 1 Foundation · Q19: Add two mixed numbers and divide a mixed number by a fraction.
Specimen · Paper 1 Foundation · Q19: Find the cheapest combination of milk bottles to cover a week, then explain how using more milk would affect the cost.
Specimen · Paper 2 Foundation · Q8: Put four fractions with different denominators in order from smallest.
Specimen · Paper 2 Foundation · Q15: Find a fraction of a length and evaluate a calculation with a square and a square root.