Surds in IGCSE Maths New from 2025
Simplifying surds, adding and subtracting, and rationalising the denominator.
0580 Extended · E1.8What is a surd?
A surd is a root that cannot be simplified to a whole number. For example, \(\sqrt{9} = 3\) is not a surd because it simplifies exactly, but \(\sqrt{5}\) is a surd because it gives an irrational decimal that never terminates or repeats.
On the IGCSE, surds are used to give exact answers rather than rounded decimals. When a question says "give your answer in surd form" or "give an exact value", this is what it means.
These two rules underpin all surd work:
\[ \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \] \[ \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \]Both require \(a \geqslant 0\) and \(b > 0\). You'll use the first rule constantly when simplifying.
Simplifying surds
To simplify a surd, find the largest square factor of the number under the root, then split:
\[ \begin{aligned} \sqrt{n} &= \sqrt{a^2 \times b} \\[6pt] &= \sqrt{a^2} \times \sqrt{b} \\[6pt] &= a\sqrt{b} \end{aligned} \]where \(a^2\) is the largest square factor. If it isn't obvious, use prime factorisation: pair up equal primes, and one from each pair comes out of the root.
(a) Simplify \(\sqrt{20}\).
(b) Simplify \(\sqrt{72}\).
Part (a)
The largest square factor of 20 is 4:
\[ \begin{aligned} \sqrt{20} &= \sqrt{4 \times 5} \\[6pt] &= \sqrt{4} \times \sqrt{5} \\[6pt] &= 2\sqrt{5} \end{aligned} \]Part (b)
The largest square factor of 72 is 36:
\[ \begin{aligned} \sqrt{72} &= \sqrt{36 \times 2} \\[6pt] &= \sqrt{36} \times \sqrt{2} \\[6pt] &= 6\sqrt{2} \end{aligned} \]If you can't spot the largest square factor, factorise fully:
\[ 72 = 2^3 \times 3^2 \]Pair up the primes: one pair of 3s comes out as a 3, one pair of 2s comes out as a 2, one 2 stays inside.
\[ \sqrt{72} = 2 \times 3 \times \sqrt{2} = 6\sqrt{2} \]Adding & subtracting surds
You can only add or subtract surds that have the same number under the root — just like collecting like terms in algebra:
\[ a\sqrt{n} + b\sqrt{n} = (a + b)\sqrt{n} \]If the surds don't match, simplify them first. They may become like surds after simplification.
Simplify \(\sqrt{200} - \sqrt{32}\).
Simplify each surd first:
\[ \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \] \[ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \]Now subtract:
\[ 10\sqrt{2} - 4\sqrt{2} = 6\sqrt{2} \]Rationalising the denominator
It is a mathematical convention that we don't leave a surd in the denominator of a fraction. Rationalising means rewriting the fraction so the denominator is a rational number (no roots).
If the denominator is \(\sqrt{n}\) or \(c\sqrt{n}\), multiply top and bottom by \(\sqrt{n}\):
\[ \begin{aligned} \frac{a}{\sqrt{n}} &= \frac{a}{\sqrt{n}} \times \frac{\sqrt{n}}{\sqrt{n}} \\[6pt] &= \frac{a\sqrt{n}}{n} \end{aligned} \]This works because \(\sqrt{n} \times \sqrt{n} = n\), which removes the root from the bottom.
Rationalise \(\dfrac{10}{\sqrt{5}}\).
Conjugate denominators
If the denominator has two terms involving a surd (e.g. \(a + \sqrt{b}\)), multiply top and bottom by the conjugate — the same expression with the opposite sign:
\[ \begin{aligned} \frac{1}{a + \sqrt{b}} \times \frac{a - \sqrt{b}}{a - \sqrt{b}} &= \frac{a - \sqrt{b}}{a^2 - b} \end{aligned} \]This uses the difference of two squares: \((a + \sqrt{b})(a - \sqrt{b}) = a^2 - b\), which eliminates the surd from the denominator.
Rationalise \(\dfrac{1}{-1 + \sqrt{3}}\).
The conjugate of \(-1 + \sqrt{3}\) is \(-1 - \sqrt{3}\). Multiply top and bottom:
\[ \begin{aligned} &\frac{1}{-1 + \sqrt{3}} \times \frac{-1 - \sqrt{3}}{-1 - \sqrt{3}} \\[6pt] &= \frac{-1 - \sqrt{3}}{(-1)^2 - (\sqrt{3})^2} \\[6pt] &= \frac{-1 - \sqrt{3}}{1 - 3} \\[6pt] &= \frac{-1 - \sqrt{3}}{-2} \end{aligned} \]Simplify by dividing through by \(-1\):
\[ \boxed{\frac{1 + \sqrt{3}}{2}} \]Rationalise \(\dfrac{5}{3 + \sqrt{2}}\).
Multiply by the conjugate \(\dfrac{3 - \sqrt{2}}{3 - \sqrt{2}}\):
\[ \begin{aligned} &\frac{5(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} \\[6pt] &= \frac{15 - 5\sqrt{2}}{9 - 2} \\[6pt] &= \boxed{\frac{15 - 5\sqrt{2}}{7}} \end{aligned} \]Surds appear most often on the non-calculator Paper 2. When a question says "give your answer in the form \(a + b\sqrt{c}\)" or "give an exact answer", it's asking you to rationalise and simplify. Don't convert to decimals.
Exam-style question
Do not use a calculator in this question. Show all your working.
(a) Simplify \(\sqrt{75}\). [1]
(b) Simplify \(\sqrt{75} + \sqrt{27}\). [2]
(c) Expand and simplify \((4 + \sqrt{3})(4 - \sqrt{3})\). [2]
(d) Rationalise the denominator of \(\dfrac{2}{4 + \sqrt{3}}\).
Give your answer in the form \(\dfrac{a - b\sqrt{3}}{c}\) where \(a\), \(b\) and \(c\) are integers. [3]
Part (a)
\[ \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \]Part (b)
Simplify \(\sqrt{27}\) first:
\[ \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \]Now add:
\[ 5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3} \]Part (c)
This is a difference of two squares:
\[ \begin{aligned} (4 + \sqrt{3})(4 - \sqrt{3}) &= 4^2 - (\sqrt{3})^2 \\[6pt] &= 16 - 3 \\[6pt] &= 13 \end{aligned} \]Part (d)
Multiply by the conjugate using the result from part (c):
\[ \begin{aligned} &\frac{2}{4 + \sqrt{3}} \times \frac{4 - \sqrt{3}}{4 - \sqrt{3}} \\[6pt] &= \frac{2(4 - \sqrt{3})}{13} \\[6pt] &= \boxed{\frac{8 - 2\sqrt{3}}{13}} \end{aligned} \]So \(a = 8\), \(b = 2\), \(c = 13\).
This is a typical 8-mark surds question. Notice how each part builds on the previous one — part (a) gives you a skill you need in part (b), and part (c) sets up the denominator calculation for part (d). Recognising that structure helps you spot what's coming and plan your working.
Common mistakes
Writing \(\sqrt{72} = 2\sqrt{18}\) instead of \(6\sqrt{2}\). The surd under the root should have no remaining square factors. Always check: can the number under the root be divided by 4, 9, 16, 25...?
\(\sqrt{2} + \sqrt{3} \neq \sqrt{5}\). You cannot add surds with different numbers under the root. They are not like terms.
To rationalise \(\frac{1}{3 + \sqrt{2}}\), multiply by \(\frac{3 - \sqrt{2}}{3 - \sqrt{2}}\), not by \(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}\). The sign in the middle must change to create the difference of two squares.
After rationalising, always check whether the resulting fraction can be simplified further. For example, \(\frac{6\sqrt{3}}{3}\) simplifies to \(2\sqrt{3}\).
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