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IGCSE · 4MA1 · Topic

Surds: past paper questions & worked solutions

Every Surds question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.

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13questions
47marks in total
24%of papers (13 of 54) include it
0/12/1easy / medium / hard

Questions, easiest first

Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.

1. Jun 2026 P1H · Q17

Medium3 marks
Jun 2026 P1H Question 17 (3 marks): Surds – 17 Show that 4 20 � � �5 2 5 � � � can be written in the form a b − where a and b are integers. (Total for Question 17 is 3 marks)
Show worked solution
  1. Solution | Step 1
    \[\sqrt{20}=2\sqrt5,\qquad(4-\sqrt{20})(5-2\sqrt5)=(4-2\sqrt5)(5-2\sqrt5).\]

    Simplify the surd first.

  2. Solution | Step 2
    \[20-8\sqrt5-10\sqrt5+20=40-18\sqrt5.\]

    Expand; the product of the two surd terms is positive 20.

  3. Solution | Step 3
    \[18\sqrt5=\sqrt{18^2\times5}=\sqrt{1620},\quad(4-\sqrt{20})(5-2\sqrt5)=40-\sqrt{1620}.\]

    Thus the required integer values are a=40 and b=1620.

Final answer

\(40-\sqrt{1620}\), so \(a=40,\ b=1620\).

2. Nov 2025 4WM1H · Q16

Medium3 marks
Nov 2025 4WM1H Question 16 (3 marks): Surds – 16 Show that 2 7 2 7 3 can be written in the form a – b where a and b are integers. Show your working clearly. (Total for Question 16 is 3 marks)
Show worked solution
  1. Simplify surd
    \[ \frac{2\sqrt7+2}{\sqrt7-3}\times\frac{\sqrt7+3}{\sqrt7+3} \]
    \[ =\frac{(2\sqrt7+2)(\sqrt7+3)}{7-9} \]
    \[ =\frac{20+8\sqrt7}{-2} \]
    \[ =-10-4\sqrt7=-10-\sqrt{112} \]
Final answer

\(-10-\sqrt{112}\).

3. Jun 2022 P1H · Q16

Medium3 marks
Jun 2022 P1H Question 16 (3 marks): Surds – 16 Without using a calculator, show that 12 2 1 2 2 32 12 5 - -( ) = + Show your working clearly. (Total for Question 16 is 3 marks)
Show worked solution
  1. Proof

    Rationalise the denominator: \[ \frac{12}{\sqrt2-1} =\frac{12(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)} \]

    \[ =12(\sqrt2+1)=12\sqrt2+12 \]

    Also, \[ (\sqrt2)^5=(\sqrt2)^4\sqrt2=4\sqrt2 \]

    So the left side is \[ 12\sqrt2+12-4\sqrt2=8\sqrt2+12 \]

    And \[ 2\sqrt{32}+12=2(4\sqrt2)+12=8\sqrt2+12 \]

    So both sides are equal.

Final answer

Shown.

4. Nov 2021 P2H · Q17

Medium3 marks
Nov 2021 P2H Question 17 (3 marks): Surds – 17 Express 8 5 1 - in the form a b + where a and b are integers. Show each stage of your working clearly. (Total for Question 17 is 3 marks)
Show worked solution
  1. Simplify surd
    \[ \frac{8}{\sqrt5-1} \]
  2. Rationalise the denominator

    Rationalise the denominator:

    \[ \frac{8}{\sqrt5-1}\times \frac{\sqrt5+1}{\sqrt5+1} =\frac{8(\sqrt5+1)}{5-1} \]
    \[ =2(\sqrt5+1)=2\sqrt5+2 \]
  3. Simplify surd

    In the form \(\sqrt a+b\),

    \[ 2\sqrt5+2=\sqrt{20}+2 \]
Final answer

\(\sqrt{20}+2\)

5. Jan 2021 P1H · Q18

Medium3 marks
Jan 2021 P1H Question 18 (3 marks): Surds – 18 Given that (8 - x )(5 + x ) = y x + 21 where x is a prime number and y is an integer, find the value of x and the value of y. Show each stage of your working clearly. x = y = (Total for Question 18 is 3 marks)
Show worked solution
  1. Simplify surd
    \[ (8-\sqrt{x})(5+\sqrt{x})=40+8\sqrt{x}-5\sqrt{x}-x \]
    \[ =40+3\sqrt{x}-x \]
  2. Simplify surd

    This is equal to \[ y\sqrt{x}+21 \]

  3. Compare the rational parts

    Compare the rational parts: \[ 40-x=21 \]

    \[ x=19 \]
  4. Compare the surd coefficients

    Compare the surd coefficients: \[ y=3 \]

Final answer

\(x=19\), \(y=3\)

6. Jan 2020 P2H · Q20

Medium3 marks
Jan 2020 P2H Question 20 (3 marks): Surds – 20 The area of a rectangle is 18 cm2 The length of the rectangle is 7 1 + ( ) cm. Without using a calculator and showing each stage of your working, find the width of the rectangle. Give your answer in the form a b + c w
Show worked solution
  1. Simplify surd
    \[ \text{width}=\frac{18}{\sqrt7+1} \]
  2. Rationalise the denominator

    Rationalise the denominator:

    \[ \frac{18}{\sqrt7+1}\times \frac{\sqrt7-1}{\sqrt7-1} =\frac{18(\sqrt7-1)}{7-1} \]
    \[ =\frac{18(\sqrt7-1)}{6}=3(\sqrt7-1) \]
    \[ =3\sqrt7-3 \]
Final answer

\(3\sqrt7-3\) cm

7. May-Nov 2020 P1H · Q19

Medium3 marks
May-Nov 2020 P1H Question 19 (3 marks): Surds – 19 Without using a calculator, rationalise the denominator of 6 3 7 - Simplify your answer. You must show each stage of your working. (Total for Question 19 is 3 marks)
Show worked solution
  1. Simplify surd
    \[ \frac{6}{3-\sqrt7} \]
  2. Rationalise the denominator

    Rationalise the denominator:

    \[ \frac{6}{3-\sqrt7}\times \frac{3+\sqrt7}{3+\sqrt7} =\frac{6(3+\sqrt7)}{9-7} \]
    \[ =\frac{6(3+\sqrt7)}{2}=3(3+\sqrt7) \]
    \[ =9+3\sqrt7 \]
Final answer

\(9+3\sqrt7\)

8. May 2025 4WM1HR · Q16

Medium4 marks
May 2025 4WM1HR Question 16 (4 marks): Surds – 16 96 2 = k (a) Find the value of k k = (1) (b) Show that 8 3 5 − can be written in the form a b + where a and b are integers. Show each stage of your working. (3) (Total for Question 16 is 4 marks)
Show worked solution
  1. (a)
    \[ \sqrt{96}=4\sqrt 6=2\sqrt{24} \]
  2. Calculate value

    So \(k=24\).

  3. (b)
    \[ \frac{8}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5} \]
    \[ =\frac{8(3+\sqrt5)}{9-5}=2(3+\sqrt5) \]
    \[ =6+2\sqrt5 \]
Final answer

\(k=24\), \(6+2\sqrt5\).

9. May 2025 4WM1H · Q17

Medium4 marks
May 2025 4WM1H Question 17 (4 marks): Surds – 17 3 3 5 k where k is an integer. (a) Find the value of k k = (1) (b) Show that 21 3 2 − can be written in the form c d + where c and d are integers. Show each stage of your working clearly. (3) (Total for Question 17
Show worked solution
  1. Simplify surd
    \[ (\sqrt3)^5=(\sqrt3)^4\sqrt3=3^2\sqrt3=9\sqrt3 \]
  2. Simplify surd

    So \(k=9\).

  3. Split the ratio

    For part (b), rationalise the denominator.

    \[ \frac{21}{3-\sqrt2}\times\frac{3+\sqrt2}{3+\sqrt2} =\frac{21(3+\sqrt2)}{9-2} \]
    \[ =3(3+\sqrt2)=9+3\sqrt2 \]
  4. Simplify surd

    Since \(3\sqrt2=\sqrt{18}\),

    \[ 9+3\sqrt2=9+\sqrt{18} \]
Final answer

\(k=9\), and \(9+\sqrt{18}\).

10. Jun 2023 P2H · Q23

Medium4 marks
Jun 2023 P2H Question 23 (4 marks): Surds – 23 The diagram shows a cuboid with a square cross section. x cm x cm (2 5 - 3) cm Diagram NOT accurately drawn The volume of the cuboid is (13 + 6 5 ) cm3 Without using a calculator, find the value of x Give your answer
Show worked solution
  1. The volume is
    \[ x^2(2\sqrt5-3)=13+6\sqrt5 \]
  2. Simplify surd

    So

    \[ x^2=\frac{13+6\sqrt5}{2\sqrt5-3} \]
  3. Rationalise

    Rationalise:

    \[ x^2=\frac{(13+6\sqrt5)(2\sqrt5+3)}{(2\sqrt5)^2-3^2} \]
    \[ =\frac{26\sqrt5+39+60+18\sqrt5}{20-9} \]
    \[ =\frac{99+44\sqrt5}{11}=9+4\sqrt5 \]
  4. Simplify surd

    Since

    \[ (\sqrt5+2)^2=5+4+4\sqrt5=9+4\sqrt5 \]
    \[ x=\sqrt5+2 \]
Final answer

\(2+\sqrt5\)

11. Jun 2022 P1HR · Q21

Medium4 marks
Jun 2022 P1HR Question 21 (4 marks): Surds – 21 Express 3 8 2 1 2 + - ( ) in the form p + q where p and q are integers. Show each stage of your working clearly. (Total for Question 21 is 4 marks)
Show worked solution
  1. Simplify surd
    \[ \frac{3+\sqrt8}{(\sqrt2-1)^2} \]
  2. Simplify surd

    Since \(\sqrt8=2\sqrt2\),

    \[ \frac{3+2\sqrt2}{(\sqrt2-1)^2} \]
  3. Simplify surd

    Also

    \[ (\sqrt2-1)^2=3-2\sqrt2 \]
  4. Simplify surd

    So

    \[ \frac{3+2\sqrt2}{3-2\sqrt2}\times \frac{3+2\sqrt2}{3+2\sqrt2} =\frac{(3+2\sqrt2)^2}{9-8} \]
    \[ =(3+2\sqrt2)^2=9+12\sqrt2+8 \]
    \[ =17+12\sqrt2=17+\sqrt{288} \]
Final answer

\(17+\sqrt{288}\)

12. Specimen 4WM1H · Q17

Medium4 marks
Specimen 4WM1H Question 17 (4 marks): Surds – 17 Express (3 + sqrt(8))/(sqrt(2) - 1)^2 in the form p + sqrt(q), where p and q are integers. (Total for Question 17 is 4 marks)
Show worked solution
  1. Simplify the surds
    \[ \sqrt8=2\sqrt2 \]
    \[ (\sqrt2-1)^2=2-2\sqrt2+1=3-2\sqrt2 \]
  2. Use the conjugate
    \[ \frac{3+2\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2} \]

    The denominator is

    \[ 9-(2\sqrt2)^2=9-8=1 \]
  3. Expand the numerator
    \[ (3+2\sqrt2)^2=9+12\sqrt2+8=17+12\sqrt2 \]
  4. Write in the requested form
    \[ 12\sqrt2=\sqrt{144\times 2}=\sqrt{288} \]
Final answer

\(17+\sqrt{288}\).

13. May 2026 4WM1H · Q16

Hard6 marks
May 2026 4WM1H Question 16 (6 marks): Surds – 16 (a) Solve x x 2 2 3 13 0 Give your solutions in the form a b ± where a and b are integers. Show your working clearly. (3) (b) Show that 20 5 2 + can be written in the form c d − where c and d are integers. (3) (Tota
Show worked solution
  1. (a) Quadratic formula
    \[x=\frac{2\sqrt3\pm\sqrt{(2\sqrt3)^2+4(13)}}{2}=\frac{2\sqrt3\pm\sqrt{64}}{2}\]
  2. Simplify
    \[x=\frac{2\sqrt3\pm8}{2}=\sqrt3\pm4\]
  3. (b) Rationalise
    \[\frac{\sqrt{20}}{\sqrt5+2}\times\frac{\sqrt5-2}{\sqrt5-2}=\frac{2\sqrt5(\sqrt5-2)}{5-4}\]
  4. Simplify
    \[=10-4\sqrt5=10-\sqrt{80}\quad(c=10,\ d=80)\]
Final answer

(a) \(\sqrt3\pm4\) (b) \(10-\sqrt{80}\)

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Dr Eslam Ahmed

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