Surds – GCSE Maths exam questions with mark scheme

1Grade 6–7HigherNon-calculator3 marks

(a) Simplify $\sqrt{75}$ [1]

(b) Simplify $\sqrt{48} + \sqrt{27}$. Give your answer in the form $a\sqrt{3}$ where $a$ is an integer. [2]

(a) $5\sqrt{3}$B1
(b) for $\sqrt{48} = 4\sqrt{3}$ or $\sqrt{27} = 3\sqrt{3}$M1
(b) $7\sqrt{3}$A1
Answer: (a) $5\sqrt{3}$ (b) $7\sqrt{3}$
Examiner tip: Look for the largest square factor (48 = 16 × 3); $\sqrt{48} + \sqrt{27} \ne \sqrt{75}$.

2Grade 6–7HigherNon-calculator2 marks

Rationalise the denominator of $\dfrac{12}{\sqrt{3}}$

Give your answer in its simplest form. [2]

for $\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$ or $\frac{12\sqrt{3}}{3}$M1
$4\sqrt{3}$A1
Answer: $4\sqrt{3}$
Examiner tip: Leaving $\frac{12\sqrt{3}}{3}$ unsimplified loses the accuracy mark.

3Grade 6–7HigherNon-calculator2 marks

Expand and simplify $(3 + \sqrt{2})(5 - \sqrt{2})$

Give your answer in the form $a + b\sqrt{2}$, where $a$ and $b$ are integers. [2]

for at least 3 of the 4 terms correct: $15 - 3\sqrt{2} + 5\sqrt{2} - 2$M1
$13 + 2\sqrt{2}$A1
Answer: $13 + 2\sqrt{2}$
Examiner tip: $\sqrt{2} \times \sqrt{2} = 2$, and watch the sign: $+\sqrt{2} \times -\sqrt{2} = -2$.

4Grade 8–9HigherNon-calculator3 marks

Show that $\dfrac{4 + \sqrt{5}}{3 - \sqrt{5}}$ can be written in the form $\dfrac{a + b\sqrt{5}}{c}$, where $a$, $b$ and $c$ are integers. [3]

for multiplying numerator and denominator by $3 + \sqrt{5}$ (or an equivalent such as $-3 - \sqrt{5}$)M1
for numerator expanded with at least 3 of the 4 terms correct, e.g. $12 + 4\sqrt{5} + 3\sqrt{5} + 5$, with denominator $9 - 5$ or $4$M1
$\frac{17 + 7\sqrt{5}}{4}$ (accept $a = 17$, $b = 7$, $c = 4$ stated; accept any correct equivalent with integers $a$, $b$, $c$, e.g. $\frac{34 + 14\sqrt{5}}{8}$, since the question does not require simplest form)A1
Answer: $\dfrac{17 + 7\sqrt{5}}{4}$
Examiner tip: Multiply by the conjugate $3 + \sqrt{5}$ (change the sign), so the denominator becomes $9 - 5 = 4$.

5Grade 8–9HigherNon-calculator4 marks

The diagram shows a right-angled triangle. The two shorter sides have lengths $(2 + \sqrt{3})$ cm and $(2 - \sqrt{3})$ cm.

(a) Show that the area of the triangle is $\dfrac{1}{2}$ cm². [2]

(b) Find the exact length, $x$ cm, of the hypotenuse. [2]

(a) for $\frac{1}{2}(2 + \sqrt{3})(2 - \sqrt{3})$ with expansion, e.g. $4 - 2\sqrt{3} + 2\sqrt{3} - 3$M1
(a) for $\frac{1}{2} \times 1 = \frac{1}{2}$ shownA1
(b) for $(2 + \sqrt{3})^2 + (2 - \sqrt{3})^2$ expanded, e.g. $7 + 4\sqrt{3} + 7 - 4\sqrt{3}$M1
(b) $\sqrt{14}$A1
Answer: (a) $\frac{1}{2}(4 - 3) = \frac{1}{2}$ (b) $x = \sqrt{14}$
Examiner tip: $(2 + \sqrt{3})^2$ is $4 + 4\sqrt{3} + 3$, not $4 + 3$ – write out both brackets.

6Grade 8–9HigherNon-calculator3 marks

$(p + \sqrt{3})^2 = 28 + q\sqrt{3}$, where $p$ and $q$ are positive integers.

Find the value of $p$ and the value of $q$. [3]

for $p^2 + p\sqrt{3} + p\sqrt{3} + 3$ or $p^2 + 2p\sqrt{3} + 3$ (also award for a correct full expansion of $(k + \sqrt{3})^2$ for a trial integer value $k$, e.g. $(5 + \sqrt{3})^2 = 25 + 10\sqrt{3} + 3$)M1
$p = 5$ (from $p^2 + 3 = 28$)A1
$q = 10$ (ft $q = 2p$ for their positive integer $p$)A1
Answer: $p = 5$, $q = 10$
Examiner tip: Expand fully ($\sqrt{3} \times \sqrt{3} = 3$), then match the whole-number parts ($p^2 + 3 = 28$) and the $\sqrt{3}$ parts ($2p = q$) separately.

7Grade 6–7HigherNon-calculator2 marks

Show that $\dfrac{\sqrt{18} + \sqrt{8}}{\sqrt{2}}$ is an integer. [2]

for $\sqrt{18} = 3\sqrt{2}$ and $\sqrt{8} = 2\sqrt{2}$, or $\sqrt{9} + \sqrt{4}$M1
for $\frac{5\sqrt{2}}{\sqrt{2}} = 5$ or $3 + 2 = 5$ shownA1
Answer: $= 5$
Examiner tip: Simplify each surd to a multiple of $\sqrt{2}$ first, then divide.

8Grade 4–5HigherNon-calculator3 marks

(a) Simplify $\sqrt{63}$ [1]

(b) Work out $\sqrt{6} \times \sqrt{10}$

Give your answer in the form $a\sqrt{b}$, where $a$ and $b$ are integers and $a > 1$. [2]

(a) $3\sqrt{7}$ caoB1
(b) $\sqrt{60}$ seen, or $\sqrt{4 \times 15}$, or $\sqrt{2}\times\sqrt{3}\times\sqrt{2}\times\sqrt{5}$ seenM1
(b) $2\sqrt{15}$ cao (do not accept $\sqrt{4}\sqrt{15}$ as final answer)A1
Answer: (a) $3\sqrt{7}$ (b) $2\sqrt{15}$
Examiner tip: In (b), $\sqrt{60}$ is correct but not in the form asked: you must take out the square factor 4 to get the A mark.

9Grade 4–5HigherNon-calculator3 marks

Here are five numbers.

$$\sqrt{25} \qquad \sqrt{20} \qquad \sqrt{3}\times\sqrt{12} \qquad \pi \qquad \frac{\sqrt{50}}{\sqrt{2}}$$

(a) Write down the two numbers that are irrational. [1]

(b) Sam says, “$\sqrt{3}\times\sqrt{12}$ must be irrational, because $\sqrt{3}$ and $\sqrt{12}$ are both irrational.”

Is Sam correct? Give a reason for your answer. [1]

(a) $\sqrt{20}$ and $\pi$ only (both needed, no extras)B1
(b) No, with reason, e.g. $\sqrt{3}\times\sqrt{12} = \sqrt{36} = 6$, which is rational / an integer (accept ‘= 6’ shown as the reason)C1
(c) any correct pair of different irrational numbers, e.g. $\sqrt{2}$ and $\sqrt{50}$, or $\sqrt{5}$ and $2\sqrt{5}$; allow $\pi$ and $\frac{10}{\pi}$; do not accept $\sqrt{10}$ and $\sqrt{10}$ (not different)B1
Answer: (a) $\sqrt{20}$, $\pi$ (b) No: $\sqrt{36}=6$ is rational (c) e.g. $\sqrt{2}$ and $\sqrt{50}$
Examiner tip: Simplify each number before deciding: $\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5$ looks irrational but is not.

10Grade 6–7HigherNon-calculator3 marks

Show that $\dfrac{6 + \sqrt{8}}{\sqrt{2}}$ can be written in the form $a + b\sqrt{2}$, where $a$ and $b$ are integers. [3]

Multiplies numerator and denominator by $\sqrt{2}$, e.g. $\dfrac{(6+\sqrt{8})\sqrt{2}}{2}$, or writes $\sqrt{8} = 2\sqrt{2}$M1
Correct expansion or division, e.g. $\dfrac{6\sqrt{2} + 4}{2}$, or $\dfrac{6}{\sqrt{2}} + 2$M1
$2 + 3\sqrt{2}$ (accept $3\sqrt{2} + 2$)A1
Answer: $2 + 3\sqrt{2}$
Examiner tip: The common slip is $\sqrt{8}\times\sqrt{2} = \sqrt{16}$ written as 8 instead of 4.

11Grade 6–7HigherNon-calculator3 marks

Expand and simplify $(2\sqrt{5} - 3)^2$

Give your answer in the form $a - b\sqrt{5}$, where $a$ and $b$ are positive integers. [3]

Writes as $(2\sqrt{5}-3)(2\sqrt{5}-3)$ and expands to 4 terms with at least 3 correct, e.g. $20 - 6\sqrt{5} - 6\sqrt{5} + 9$M1
$2\sqrt{5} \times 2\sqrt{5} = 20$ seen (not 10 or $4\sqrt{25}$ left unsimplified)M1
$29 - 12\sqrt{5}$ caoA1
Answer: $29 - 12\sqrt{5}$
Examiner tip: $(2\sqrt{5})^2$ is $4 \times 5 = 20$, not 10 and not $2 \times 5$; squaring a bracket also needs the two middle terms.

12Grade 6–7HigherNon-calculator4 marks

The diagram shows a rectangle.

The length of the rectangle is $(5 + \sqrt{2})$ cm. The width of the rectangle is $(\sqrt{8} - 1)$ cm.

(a) Show that the perimeter of the rectangle is $(8 + 6\sqrt{2})$ cm. [2]

(b) Work out the area of the rectangle. Give your answer in the form $(a + b\sqrt{2})$ cm², where $a$ and $b$ are integers. [2]

(a) $\sqrt{8} = 2\sqrt{2}$ seen, or $2(5+\sqrt{2}) + 2(\sqrt{8}-1)$ seenM1
(a) $10 + 2\sqrt{2} + 4\sqrt{2} - 2 = 8 + 6\sqrt{2}$ with all steps shownA1
(b) $(5+\sqrt{2})(2\sqrt{2}-1)$ expanded to 4 terms with at least 3 correct, e.g. $10\sqrt{2} - 5 + 4 - \sqrt{2}$M1
(b) $-1 + 9\sqrt{2}$ (accept $9\sqrt{2} - 1$)A1
Answer: (a) $8 + 6\sqrt{2}$ (b) $(-1 + 9\sqrt{2})$ cm²
Examiner tip: Simplify $\sqrt{8}$ to $2\sqrt{2}$ first so the surd terms can be collected; $\sqrt{2}\times 2\sqrt{2} = 4$, not $2\sqrt{4}$ left unsimplified.

13Grade 8–9HigherNon-calculator4 marks

Solve $x\sqrt{3} - 2 = x + 4$

Give your answer in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers. [4]

Collects $x$ terms on one side, e.g. $x\sqrt{3} - x = 6$M1
Factorises, $x(\sqrt{3} - 1) = 6$, so $x = \dfrac{6}{\sqrt{3}-1}$M1
Multiplies numerator and denominator by $\sqrt{3} + 1$, e.g. $\dfrac{6(\sqrt{3}+1)}{3 - 1}$M1
$x = 3 + 3\sqrt{3}$ caoA1
Answer: $x = 3 + 3\sqrt{3}$
Examiner tip: To get $x$ alone you must factorise $x(\sqrt{3}-1)$ and then rationalise using $\sqrt{3}+1$; multiplying by $\sqrt{3}$ does not clear the denominator.

14Grade 6–7HigherNon-calculator3 marks

Here are the first three terms of a geometric sequence.

$$\sqrt{3} \qquad 3 \qquad 3\sqrt{3}$$

(a) Write down the 4th term of the sequence. [1]

(b) Find the first term of the sequence that is greater than 200. You must show your working. [2]

(a) 9B1
(b) continues the sequence multiplying by $\sqrt{3}$ at least to $81\sqrt{3}$, with a reason that $81\sqrt{3} < 200$ (e.g. $\sqrt{3} < 2$ so $81\sqrt{3} < 162$, or $81\sqrt{3} = \sqrt{19683} < \sqrt{40000}$)M1
(b) 243 (accept ‘the 10th term, 243’)A1
Answer: (a) 9 (b) 243
Examiner tip: Each term is the previous one times $\sqrt{3}$, so every second term is a whole number; justify that $81\sqrt{3}$ is below 200 rather than just guessing.

15Grade 8–9HigherNon-calculator4 marks

Show that

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

is an integer. [4]

Rationalises one fraction correctly, e.g. $\dfrac{1}{\sqrt{2}+1}\times\dfrac{\sqrt{2}-1}{\sqrt{2}-1} = \dfrac{\sqrt{2}-1}{1}$M1
Rationalises all three, with each denominator simplified to 1: $\sqrt{2}-1$, $\sqrt{3}-\sqrt{2}$, $2-\sqrt{3}$M1
Adds the three results, e.g. $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3})$M1
$= 1$, with all surd terms shown to cancelA1
Answer: $1$
Examiner tip: Multiply each fraction by the conjugate of its own denominator; each denominator becomes a difference of two squares equal to 1.

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