TMUA: Indices and surds

Indices and surds turn up everywhere in the TMUA, usually as a quick simplification inside a longer question. Without a calculator you need exact rules, not decimals.

Section 1 Part 1 · MM1 Algebra and functions · spec MM1.1-MM1.2
Papers 1 and 2
8 practice questions
No calculator

Practise indices and surds →Timed TMUA paper

What the specification covers

Laws of indices for rational exponents
Simplifying surds; rationalising denominators including a + b√c forms
Comparing sizes of surd and index expressions without a calculator

Key ideas

\(a^m\times a^n=a^{m+n}\), \(a^m\div a^n=a^{m-n}\), \((a^m)^n=a^{mn}\), \(a^{-n}=\dfrac1{a^n}\), \(a^{p/q}=\left(\sqrt[q]{a}\right)^p\).
Simplify a surd by taking out square factors: \(\sqrt{72}=\sqrt{36\times2}=6\sqrt2\).
Rationalise with the conjugate: \(\dfrac{1}{\sqrt a-\sqrt b}=\dfrac{\sqrt a+\sqrt b}{a-b}\), because \((\sqrt a-\sqrt b)(\sqrt a+\sqrt b)=a-b\).
To compare numbers like \(2^{1/2}\) and \(3^{1/3}\), raise both to a common power (here 6): \(8<9\), so \(3^{1/3}\) is larger.

Common mistakes

\(\sqrt{a+b}\ne\sqrt a+\sqrt b\).
\((-8)^{2/3}=4\) but fractional powers of negative numbers need care; the TMUA keeps to positive bases.

Exam tip

When options are close, square both or raise to a common power rather than estimating decimals.

Worked example

Which of the following is equal to \(\dfrac{\sqrt{12}+\sqrt{3}}{\sqrt{3}-1}\)?

\(\frac{9-3\sqrt{3}}{2}\)
\(\frac{3+3\sqrt{3}}{2}\)
\(3\sqrt{3}+3\)
\(\frac{9+3\sqrt{3}}{2}\)
\(9+3\sqrt{3}\)

Answer: D: \(\frac{9+3\sqrt{3}}{2}\)

\(\sqrt{12}=2\sqrt{3}\), so the numerator is \(3\sqrt{3}\). Multiply top and bottom by \(\sqrt{3}+1\): \(\dfrac{3\sqrt{3}(\sqrt{3}+1)}{3-1}=\dfrac{9+3\sqrt{3}}{2}\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

Which of the following numbers is the largest?

\(6^{1/6}\)
\(2^{1/2}\)
\(8^{1/8}\)
\(5^{1/5}\)
\(3^{1/3}\)

Show the answer and solution

Answer: E: \(3^{1/3}\)

Raise pairs to a common power. \((3^{1/3})^6=9>8=(2^{1/2})^6\), so \(3^{1/3}>2^{1/2}\). \((2^{1/2})^{10}=32>25=(5^{1/5})^{10}\). \((2^{1/2})^{6}=8>6=(6^{1/6})^6\). \(8^{1/8}=2^{3/8}<2^{1/2}\). So \(3^{1/3}\) is the largest.

Question 2

Which of the following is equal to \(\dfrac{12}{\sqrt{6}-\sqrt{3}}\)?

\(12 \sqrt{3} + 12 \sqrt{6}\)
\(4 \sqrt{3} + 4 \sqrt{6}\)
\(- 4 \sqrt{3} + 4 \sqrt{6}\)
\(- 12 \sqrt{3} + 12 \sqrt{6}\)
\(\frac{4 \sqrt{3}}{3} + \frac{4 \sqrt{6}}{3}\)

Show the answer and solution

Answer: B: \(4 \sqrt{3} + 4 \sqrt{6}\)

Multiply top and bottom by \(\sqrt{6}+\sqrt{3}\): the denominator becomes \(6-3=3\), so the value is \(\dfrac{12(\sqrt{6}+\sqrt{3})}{3}\) \(4 \sqrt{3} + 4 \sqrt{6}\).

Question 3

Which of the following is equal to \(\dfrac{4}{\sqrt{6}-\sqrt{3}}\)?

\(\frac{- 4 \sqrt{3} + 4 \sqrt{6}}{3}\)
\(\frac{4 \sqrt{3} + 4 \sqrt{6}}{3}\)
\(\frac{4 \sqrt{3}}{9} + \frac{4 \sqrt{6}}{9}\)
\(- 4 \sqrt{3} + 4 \sqrt{6}\)
\(4 \sqrt{3} + 4 \sqrt{6}\)

Show the answer and solution

Answer: B: \(\frac{4 \sqrt{3} + 4 \sqrt{6}}{3}\)

Multiply top and bottom by \(\sqrt{6}+\sqrt{3}\): the denominator becomes \(6-3=3\), so the value is \(\dfrac{4(\sqrt{6}+\sqrt{3})}{3}\) \(\frac{4 \sqrt{3} + 4 \sqrt{6}}{3}\).

Question 4

Which of the following is equal to \(\dfrac{3}{\sqrt{11}-\sqrt{3}}\)?

\(\frac{3 \sqrt{3}}{14} + \frac{3 \sqrt{11}}{14}\)
\(3 \sqrt{3} + 3 \sqrt{11}\)
\(- 3 \sqrt{3} + 3 \sqrt{11}\)
\(\frac{3 \sqrt{3} + 3 \sqrt{11}}{8}\)
\(\frac{- 3 \sqrt{3} + 3 \sqrt{11}}{8}\)

Show the answer and solution

Answer: D: \(\frac{3 \sqrt{3} + 3 \sqrt{11}}{8}\)

Multiply top and bottom by \(\sqrt{11}+\sqrt{3}\): the denominator becomes \(11-3=8\), so the value is \(\dfrac{3(\sqrt{11}+\sqrt{3})}{8}\) \(\frac{3 \sqrt{3} + 3 \sqrt{11}}{8}\).

Question 5

Which of the following is equal to \(\dfrac{10}{\sqrt{13}-\sqrt{3}}\)?

\(\sqrt{3} + \sqrt{13}\)
\(- \sqrt{3} + \sqrt{13}\)
\(\frac{5 \sqrt{3}}{8} + \frac{5 \sqrt{13}}{8}\)
\(10 \sqrt{3} + 10 \sqrt{13}\)
\(- 10 \sqrt{3} + 10 \sqrt{13}\)

Show the answer and solution

Answer: A: \(\sqrt{3} + \sqrt{13}\)

Multiply top and bottom by \(\sqrt{13}+\sqrt{3}\): the denominator becomes \(13-3=10\), so the value is \(\dfrac{10(\sqrt{13}+\sqrt{3})}{10}\) \(\sqrt{3} + \sqrt{13}\).

Question 6

Which of the following is equal to \(\dfrac{10}{\sqrt{10}-\sqrt{3}}\)?

\(\frac{10 \sqrt{3} + 10 \sqrt{10}}{7}\)
\(\frac{- 10 \sqrt{3} + 10 \sqrt{10}}{7}\)
\(\frac{10 \sqrt{3}}{13} + \frac{10 \sqrt{10}}{13}\)
\(- 10 \sqrt{3} + 10 \sqrt{10}\)
\(10 \sqrt{3} + 10 \sqrt{10}\)

Show the answer and solution

Answer: A: \(\frac{10 \sqrt{3} + 10 \sqrt{10}}{7}\)

Multiply top and bottom by \(\sqrt{10}+\sqrt{3}\): the denominator becomes \(10-3=7\), so the value is \(\dfrac{10(\sqrt{10}+\sqrt{3})}{7}\) \(\frac{10 \sqrt{3} + 10 \sqrt{10}}{7}\).

Question 7

Which of the following is equal to \(\dfrac{12}{\sqrt{13}-\sqrt{3}}\)?

\(\frac{- 6 \sqrt{3} + 6 \sqrt{13}}{5}\)
\(\frac{3 \sqrt{3}}{4} + \frac{3 \sqrt{13}}{4}\)
\(\frac{6 \sqrt{3} + 6 \sqrt{13}}{5}\)
\(12 \sqrt{3} + 12 \sqrt{13}\)
\(- 12 \sqrt{3} + 12 \sqrt{13}\)

Show the answer and solution

Answer: C: \(\frac{6 \sqrt{3} + 6 \sqrt{13}}{5}\)

Multiply top and bottom by \(\sqrt{13}+\sqrt{3}\): the denominator becomes \(13-3=10\), so the value is \(\dfrac{12(\sqrt{13}+\sqrt{3})}{10}\) \(\frac{6 \sqrt{3} + 6 \sqrt{13}}{5}\).

Keep going

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