Surds (Higher)

This is a Higher-tier [H] lesson throughout — every section below is Higher-only content in OCR GCSE Mathematics (J560). A surd is a root that cannot be simplified to a rational number, such as $\sqrt{2}$ or $\sqrt{12}$ . Surds let us give exact answers rather than rounded decimals, which is why "leave your answer in surd form" is a common instruction on the Higher papers. This lesson covers simplifying surds, the rules for surd arithmetic (multiplying, dividing, adding and subtracting), and rationalising a denominator.

This lesson mainly builds AO1 fluency in manipulating surds, with AO2 reasoning when you justify that a number is irrational, and AO3 problem-solving where surds arise from areas, Pythagoras or expanding brackets.

Key Vocabulary

Term	Meaning
Surd	A root that is irrational, e.g. $\sqrt{2}$ , $\sqrt{12}$ , $\sqrt[3]{5}$
Rational number	A number expressible as a fraction of integers
Irrational number	A number that cannot be written as such a fraction
Simplify a surd	Write it with the smallest possible number under the root
Rationalise	Remove a surd from the denominator of a fraction
Like surds	Surds with the same number under the root, e.g. $3\sqrt{2}$ and $5\sqrt{2}$

Why Surds Matter

$\sqrt{2} = 1.41421356\ldots$ never terminates and never recurs, so any decimal we write down is only an approximation. Leaving the answer as $\sqrt{2}$ keeps it exact. The two rules that drive all surd work are:

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}, \qquad \dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}.$

Note that there is no such rule for addition: $\sqrt{a} + \sqrt{b} \ne \sqrt{a+b}$ in general.

A quick way to see which roots are surds: a square root is a surd exactly when the number under it is not a perfect square. So $\sqrt{9} = 3$ is not a surd (it is a whole number), but $\sqrt{10}$ is a surd because $10$ lies between the perfect squares $9$ and $16$ and has no exact square root. The same idea works for cube roots and perfect cubes. Recognising this stops you wasting time trying to "simplify" a root that is already rational, and tells you immediately when an answer should be left in surd form.

To check the addition rule fails, test it with numbers: $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ , but $\sqrt{9+16} = \sqrt{25} = 5$ . Since $7 \ne 5$ , there is clearly no "add under the root" rule — a useful sanity check whenever you are tempted to combine surds incorrectly.

Simplifying Surds

To simplify a surd, find the largest square factor of the number under the root, then split it off.

Worked Example 1

Simplify $\sqrt{12}$ .

The largest square factor of $12$ is $4$ ( $12 = 4 \times 3$ ).

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}.$

Answer: $2\sqrt{3}$ .

Worked Example 2

Simplify $\sqrt{72}$ .

The largest square factor of $72$ is $36$ ( $72 = 36 \times 2$ ).

$\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}.$

Answer: $6\sqrt{2}$ .

Common error: Using a smaller square factor, e.g. $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$ . This is true but not fully simplified — $\sqrt{18}$ simplifies again. Always take the largest square factor.

Worked Example 2b

Write $5\sqrt{3}$ as a single square root.

This is the reverse of simplifying: take the number outside back inside by squaring it. $5 = \sqrt{25}$ , so $5\sqrt{3} = \sqrt{25} \times \sqrt{3} = \sqrt{25 \times 3} = \sqrt{75}$ .

Answer: $\sqrt{75}$ . This reverse skill is handy when you need to compare two surds — for example, is $5\sqrt{3}$ bigger than $4\sqrt{5}$ ? Writing both as single roots, $\sqrt{75}$ and $\sqrt{80}$ , shows at once that $4\sqrt{5} = \sqrt{80}$ is the larger.

Multiplying and Dividing Surds

Use $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ and simplify the result; multiply any whole-number parts separately.

Worked Example 3

Work out $\sqrt{6} \times \sqrt{8}$ , giving your answer in simplest surd form.

$\sqrt{6} \times \sqrt{8} = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}.$

Answer: $4\sqrt{3}$ .

Worked Example 4

Work out $3\sqrt{5} \times 2\sqrt{10}$ .

Multiply whole numbers and surds separately: $(3 \times 2)(\sqrt{5} \times \sqrt{10}) = 6\sqrt{50} = 6 \times 5\sqrt{2} = 30\sqrt{2}$ .

Answer: $30\sqrt{2}$ .

Worked Example 5

Work out $\dfrac{\sqrt{40}}{\sqrt{5}}$ .

$\dfrac{\sqrt{40}}{\sqrt{5}} = \sqrt{\dfrac{40}{5}} = \sqrt{8} = 2\sqrt{2}.$

Answer: $2\sqrt{2}$ .

Adding and Subtracting Surds

You can only add or subtract like surds (same number under the root) — treat the surd like an algebraic term. Simplify each surd first; that often turns unlike-looking surds into like ones.

Worked Example 6

Simplify $\sqrt{50} + \sqrt{18}$ .

Simplify each: $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$ . Now they are like surds:

$5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}.$

Answer: $8\sqrt{2}$ .

Common error: Writing $\sqrt{50} + \sqrt{18} = \sqrt{68}$ . There is no "add under the root" rule — you must simplify to like surds and add the coefficients.

Rationalising the Denominator

It is conventional to write fractions with no surd in the denominator. For a single surd, multiply top and bottom by that surd.

Worked Example 7

Rationalise $\dfrac{6}{\sqrt{3}}$ .

Multiply numerator and denominator by $\sqrt{3}$ :

$\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}.$

Answer: $2\sqrt{3}$ .

Why does this work? Multiplying the denominator $\sqrt{3}$ by itself gives $\sqrt{3} \times \sqrt{3} = \sqrt{9} = 3$ , a rational number, which clears the surd from the bottom. We multiply the top by the same $\sqrt{3}$ so that we are really multiplying the whole fraction by $\dfrac{\sqrt{3}}{\sqrt{3}} = 1$ , leaving its value unchanged.

Worked Example 7b

Rationalise $\dfrac{\sqrt{5}}{\sqrt{8}}$ , giving your answer in simplest form.

Multiply top and bottom by $\sqrt{8}$ : $\dfrac{\sqrt{5} \times \sqrt{8}}{8} = \dfrac{\sqrt{40}}{8}$ . Now simplify the surd: $\sqrt{40} = 2\sqrt{10}$ , so the fraction is $\dfrac{2\sqrt{10}}{8} = \dfrac{\sqrt{10}}{4}$ .

Answer: $\dfrac{\sqrt{10}}{4}$ .

Common error: Forgetting to simplify $\sqrt{40}$ and to cancel the $\dfrac{2}{8}$ down to $\dfrac{1}{4}$ . "Simplest form" requires both steps.

Extended Worked Examples

Worked Example 8: Expanding a single bracket with surds

Expand and simplify $\sqrt{3}\left(\sqrt{6} + 2\sqrt{3}\right)$ .

Multiply through: $\sqrt{3} \times \sqrt{6} = \sqrt{18} = 3\sqrt{2}$ , and $\sqrt{3} \times 2\sqrt{3} = 2 \times 3 = 6$ .

$\sqrt{3}\left(\sqrt{6} + 2\sqrt{3}\right) = 3\sqrt{2} + 6.$

Answer: $3\sqrt{2} + 6$ .

Worked Example 9: Expanding two brackets (difference of two squares)

Expand and simplify $\left(\sqrt{5} + 2\right)\left(\sqrt{5} - 2\right)$ .

This is $(a+b)(a-b) = a^{2} - b^{2}$ with $a = \sqrt{5}$ , $b = 2$ :

$(\sqrt{5})^{2} - 2^{2} = 5 - 4 = 1.$

Answer: $1$ (a rational number — the surds cancel).

This is the idea behind rationalising a denominator that is a sum containing a surd: multiply by the conjugate.

Worked Example 10: Rationalising with a conjugate

Rationalise $\dfrac{4}{3 + \sqrt{5}}$ .

Multiply top and bottom by the conjugate $3 - \sqrt{5}$ :

$\dfrac{4}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{5}} = \dfrac{4(3 - \sqrt{5})}{(3)^{2} - (\sqrt{5})^{2}} = \dfrac{12 - 4\sqrt{5}}{9 - 5} = \dfrac{12 - 4\sqrt{5}}{4} = 3 - \sqrt{5}.$

Answer: $3 - \sqrt{5}$ .

Worked Example 11: Surds from Pythagoras

A right-angled triangle has legs of length $2$ cm and $\sqrt{5}$ cm. Work out the exact length of the hypotenuse.

By Pythagoras, $h^{2} = 2^{2} + (\sqrt{5})^{2} = 4 + 5 = 9$ , so $h = \sqrt{9} = 3$ cm.

Answer: $3$ cm (here the surd disappears).

Worked Example 11b: Mixed surd arithmetic

Simplify $\dfrac{\sqrt{18} + \sqrt{50}}{\sqrt{2}}$ , giving your answer as an integer.

Simplify the top first: $\sqrt{18} = 3\sqrt{2}$ and $\sqrt{50} = 5\sqrt{2}$ , so the numerator is $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$ .

$\dfrac{8\sqrt{2}}{\sqrt{2}} = 8.$

Answer: $8$ . The $\sqrt{2}$ in the numerator and denominator cancel, leaving a whole number — a neat reminder that surds often simplify away entirely.

Worked Example 11c: Area giving a surd answer

A square has area $48$ cm². Work out the exact length of one side, in simplest surd form.

The side is $\sqrt{48}$ . Simplify: $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ .

Answer: $4\sqrt{3}$ cm. Leaving it as $4\sqrt{3}$ is exact; writing $6.93$ cm would only be an approximation and could lose an "exact value" mark.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: Simplify fully $\sqrt{75} - \sqrt{27}$ .

Grades 3–4 response: $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$ , so $5\sqrt{3} - 3\sqrt{3} = 2\sqrt{3}$ .

Surds (Higher)

Surds (Higher)

Key Vocabulary

Why Surds Matter

Simplifying Surds

Worked Example 1

Worked Example 2

Worked Example 2b

Multiplying and Dividing Surds

Worked Example 3

Worked Example 4

Worked Example 5

Adding and Subtracting Surds

Worked Example 6

Rationalising the Denominator

Worked Example 7

Worked Example 7b

Extended Worked Examples

Worked Example 8: Expanding a single bracket with surds

Worked Example 9: Expanding two brackets (difference of two squares)

Worked Example 10: Rationalising with a conjugate

Worked Example 11: Surds from Pythagoras

Worked Example 11b: Mixed surd arithmetic

Worked Example 11c: Area giving a surd answer

Answering at different grade levels

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