Surds & index laws (fractional and negative powers) – IGCSE Maths exam questions with mark scheme

1Grade 1–3Non-calculator3 marks

Simplify

(a) $p^3 \times p^8$ [1]

(b) $\dfrac{q^{12}}{q^3}$ [1]

(a) $p^{11}$B1
(b) $q^{9}$ (do not accept $q^4$)B1
(c) $r^{20}$ (do not accept $r^9$)B1
Answer: (a) $p^{11}$ (b) $q^9$ (c) $r^{20}$
Examiner tip: Dividing powers means subtract the indices, not divide them, and a bracket raised to a power means multiply the indices: $q^4$ and $r^9$ score nothing.

2Grade 4–5Non-calculator5 marks

(a) Write down the value of $9^0$ [1]

(b) Find the value of $4^{-2}$ [1]

(d) Simplify fully $\dfrac{3a^5 \times 4a^{-2}}{6a}$ [2]

(a) $1$B1
(b) $\dfrac{1}{16}$ oe (accept $0.0625$)B1
(c) $-5$B1
(d) for a correct numerical coefficient of $2$ or a correct power of $a^2$ in an otherwise simplified expression, e.g. $2a^k$ or $ka^2$; or for $\dfrac{12a^3}{6a}$B1
(d) $2a^2$ caoB1
Answer: (a) $1$ (b) $\dfrac{1}{16}$ (c) $x=-5$ (d) $2a^2$
Examiner tip: $4^{-2}$ is $\frac{1}{16}$, not $-16$ or $-8$: a negative power gives a reciprocal, never a negative number.

3Grade 4–5HigherNon-calculator4 marks

Find the value of

(a) $64^{\frac{1}{3}}$ [1]

(b) $36^{-\frac{1}{2}}$ [1]

(a) $4$B1
(b) $\dfrac{1}{6}$ oeB1
(c) for $\sqrt[3]{125} = 5$ seen, or $125^2 = 15625$ with intention to cube root, or $(5^3)^{\frac{2}{3}}$M1
(c) $25$ caoA1
Answer: (a) $4$ (b) $\dfrac{1}{6}$ (c) $25$
Examiner tip: Take the root (the denominator) first, then the power: $\sqrt[3]{125}=5$ then $5^2=25$ is far easier than finding $\sqrt[3]{15625}$.

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4Grade 6–7HigherNon-calculator4 marks

(a) Simplify $\sqrt{75} + 2\sqrt{12}$
Give your answer in the form $k\sqrt{3}$, where $k$ is an integer. [2]

(b) Expand and simplify $(4 - \sqrt{3})^2$
Give your answer in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers. [2]

(a) for $\sqrt{75} = 5\sqrt{3}$ or $2\sqrt{12} = 4\sqrt{3}$ (or $\sqrt{12}=2\sqrt{3}$)M1
(a) $9\sqrt{3}$ caoA1
(b) for 3 out of 4 terms correct in the expansion of $(4-\sqrt{3})(4-\sqrt{3})$, e.g. $16 - 4\sqrt{3} - 4\sqrt{3} + 3$M1
(b) $19 - 8\sqrt{3}$ caoA1
Answer: (a) $9\sqrt{3}$ (b) $19 - 8\sqrt{3}$
Examiner tip: $(4-\sqrt{3})^2$ is not $16 - 3$ or $16+3$: write out all four terms; $(-\sqrt{3})\times(-\sqrt{3}) = +3$.

5Grade 8–9HigherNon-calculator5 marks

(a) Rationalise the denominator of $\dfrac{18}{\sqrt{6}}$
Give your answer in its simplest form. [2]

(b) Show that $\dfrac{\sqrt{5} + 1}{\sqrt{5} - 2}$ can be written in the form $a + b\sqrt{5}$, where $a$ and $b$ are integers.
Show each stage of your working. [3]

(a) for $\dfrac{18\sqrt{6}}{6}$ or $\dfrac{18}{\sqrt{6}}\times\dfrac{\sqrt{6}}{\sqrt{6}}$M1
(a) $3\sqrt{6}$ caoA1
(b) for multiplying numerator and denominator by $\sqrt{5}+2$, i.e. $\dfrac{(\sqrt{5}+1)(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}$M1
(b) for a correct expansion of either the numerator ($5 + 2\sqrt{5} + \sqrt{5} + 2$ or $7 + 3\sqrt{5}$) or the denominator ($5 - 4$ or $1$)M1
(b) $7 + 3\sqrt{5}$ from fully correct working with both numerator and denominator shownA1
Answer: (a) $3\sqrt{6}$ (b) $7 + 3\sqrt{5}$ ($a = 7$, $b = 3$)
Examiner tip: Multiply top AND bottom by the conjugate $\sqrt{5}+2$; multiplying by $\sqrt{5}-2$ or by $\sqrt{5}$ does not clear the surd from the denominator.

6Grade 8–9HigherCalculator5 marks

The diagram shows a right-angled triangle $ABC$.

$AB = (3 - \sqrt{2})$ cm and $BC = (3 + \sqrt{2})$ cm.

(a) Show that the length of $AC$ is $\sqrt{22}$ cm. [3]

(b) Find the area of triangle $ABC$.
Give your answer as a fraction in its simplest form. [2]

(a) for $(3-\sqrt{2})^2 + (3+\sqrt{2})^2$ (Pythagoras set up with surd lengths)M1
(a) for a correct expansion of at least one bracket, e.g. $9 - 6\sqrt{2} + 2$ or $11 + 6\sqrt{2}$M1
(a) for $11 - 6\sqrt{2} + 11 + 6\sqrt{2} = 22$ so $AC = \sqrt{22}$, with all working correct (no decimal working)A1
(b) for $\dfrac{1}{2}(3-\sqrt{2})(3+\sqrt{2})$ and an attempt to expand, e.g. $\dfrac{1}{2}(9 - 2)$M1
(b) $\dfrac{7}{2}$ cm$^2$ (accept $3\frac{1}{2}$)A1
Answer: (a) $AC^2 = 22$ so $AC = \sqrt{22}$ cm (b) $\dfrac{7}{2}$ cm$^2$
Examiner tip: In a 'show that' with surds, a decimal check (4.69...) scores nothing: the $\pm 6\sqrt{2}$ terms must be seen cancelling.

7Grade 8–9HigherNon-calculator5 marks

(a) Write $\dfrac{1}{\sqrt{8}}$ as a power of 2 [2]

(b) Solve $\dfrac{25^x}{5^{x+1}} = 125$
Show clear algebraic working. [3]

(a) for $\sqrt{8} = 2^{\frac{3}{2}}$ or $8 = 2^3$ and $\dfrac{1}{\sqrt{8}} = 8^{-\frac{1}{2}}$M1
(a) $2^{-\frac{3}{2}}$ oe, e.g. $2^{-1.5}$A1
(b) for writing all terms as powers of 5, e.g. $\dfrac{5^{2x}}{5^{x+1}} = 5^3$M1
(b) for $2x - (x+1) = 3$ oeM1
(b) $x = 4$ caoA1
Answer: (a) $2^{-\frac{3}{2}}$ (b) $x = 4$
Examiner tip: Write everything with the same base before comparing indices; $25^x = 5^{2x}$, not $5^{x+2}$.

8Grade 6–7HigherCalculator5 marks

(a) Simplify fully $(16a^8)^{\frac{1}{2}}$ [2]

(b) Simplify fully $\left(\dfrac{27x^6}{y^3}\right)^{-\frac{2}{3}}$ [3]

(a) for a correct coefficient $4$ or a correct power $a^4$, e.g. $4a^k$ or $ka^4$B1
(a) $4a^4$ caoB1
(b) for inverting the fraction, e.g. $\left(\dfrac{y^3}{27x^6}\right)^{\frac{2}{3}}$, or for evaluating $27^{\frac{2}{3}} = 9$ (or $27^{-\frac{2}{3}} = \frac{1}{9}$) seenM1
(b) for any two of $\dfrac{1}{9}$, $x^{-4}$ (or $\frac{1}{x^4}$), $y^2$ correct in a single expressionM1
(b) $\dfrac{y^2}{9x^4}$ oe, e.g. $\frac{1}{9}x^{-4}y^2$A1
Answer: (a) $4a^4$ (b) $\dfrac{y^2}{9x^4}$
Examiner tip: The power applies to the number as well as the letters: $27^{\frac{2}{3}} = 9$, and the negative sign flips the whole fraction.

9Grade 6–7HigherNon-calculator4 marks

(a) Kai writes $$8^{-\frac{1}{3}} = -2$$ Kai is wrong. Explain the mistake Kai has made and write down the correct value of $8^{-\frac{1}{3}}$ [2]

(b) Ella says, “When you multiply two irrational numbers together, the answer is always irrational.”
Use $\sqrt{2}$ and $\sqrt{18}$ to show that Ella is wrong. [2]

(a) for an explanation that a negative index means take the reciprocal (one over), not make the answer negativeC1
(a) $\dfrac{1}{2}$ oe (accept $0.5$)B1
(b) for $\sqrt{2}\times\sqrt{18} = \sqrt{36}$ or $\sqrt{2}\times 3\sqrt{2}$M1
(b) for $= 6$ with a statement that 6 is rational (an integer / can be written as $\frac{6}{1}$), so Ella is wrongC1
Answer: (a) The negative power means a reciprocal, not a negative answer; $8^{-\frac{1}{3}} = \dfrac{1}{2}$ (b) $\sqrt{2}\times\sqrt{18} = \sqrt{36} = 6$, which is rational
Examiner tip: For the final C1 you must state the conclusion: finding 6 is not enough without saying 6 is rational.

10Grade 6–7HigherNon-calculator3 marks

Here are five numbers.

$8^{\frac{2}{3}}$

$16^{-\frac{1}{2}}$

$4^0$

$\sqrt{3}\times\sqrt{12}$

$\left(\dfrac{1}{2}\right)^{-3}$

Write these numbers in order of size. Start with the smallest number.
You must show how you get your answer. [3]

for evaluating at least three of the numbers correctly: $8^{\frac{2}{3}}=4$, $16^{-\frac{1}{2}}=\frac{1}{4}$, $4^0=1$, $\sqrt{3}\times\sqrt{12}=6$, $\left(\frac{1}{2}\right)^{-3}=8$M1
for evaluating all five numbers correctlyM1
$16^{-\frac{1}{2}},\ 4^0,\ 8^{\frac{2}{3}},\ \sqrt{3}\times\sqrt{12},\ \left(\frac{1}{2}\right)^{-3}$ (accept the values $\frac{1}{4}, 1, 4, 6, 8$ given in order only if each value is linked to its original number)A1
Answer: $16^{-\frac{1}{2}}$, $4^0$, $8^{\frac{2}{3}}$, $\sqrt{3}\times\sqrt{12}$, $\left(\frac{1}{2}\right)^{-3}$ (values $\frac{1}{4}, 1, 4, 6, 8$)
Examiner tip: $\left(\frac{1}{2}\right)^{-3}$ is $2^3 = 8$, the largest; students who treat it as $-\frac{1}{8}$ put it first and lose the final mark.

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