What is not taking the largest square factor?

sqrt(72) = 2sqrt(18) is not fully simplified; pull out 36 to get 6sqrt(2).

What is wrong conjugate?

The conjugate of 3 + sqrt(5) is 3 - sqrt(5), changing only the sign between the terms.

Rationalise 12 - sqrt(3). [3 marks]

SingaporeAdditional MathematicsSyllabus dot point

How do we simplify surds and rewrite a fraction so that there is no surd in the denominator?

Simplify surds, perform the four operations with surds, and rationalise denominators of the form a over root b and a over (b plus root c)

A focused answer to the N(A)-Level Additional Mathematics outcome on surds. Simplify surds, add, subtract, multiply and divide them, and rationalise denominators including those with a conjugate.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to work confidently with surds, which are roots such as $\sqrt{2}$ or $\sqrt{3}$ that do not simplify to whole numbers. You should be able to simplify a surd by pulling out perfect-square factors, add and subtract like surds, multiply and divide them, and rationalise a denominator so that no surd is left on the bottom of a fraction. Exact surd answers are expected wherever a question asks for an exact value rather than a decimal.

The answer

What a surd is

A surd is an irrational root that is left in root form because writing it as a decimal would only be an approximation. The exact value $\sqrt{2}$ is preferred to $1.414\ldots$ The two rules that drive everything are:

\sqrt{a} \times \sqrt{b} = \sqrt{ab}, \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}

Simplifying a surd

To simplify, find the largest perfect-square factor of the number and take its root outside:

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

A surd is fully simplified when the number under the root has no perfect-square factor other than $1$ .

Adding and subtracting

You can only add or subtract like surds (the same root), exactly as you collect like terms in algebra. Simplify first so that hidden like surds appear:

\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}

Multiplying

Multiply the whole-number parts together and the surd parts together, then simplify:

3\sqrt{2} \times 4\sqrt{6} = 12\sqrt{12} = 12 \times 2\sqrt{3} = 24\sqrt{3}

Rationalising a single-surd denominator

To clear a surd such as $\sqrt{b}$ from the denominator, multiply the top and bottom by that surd:

\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}}\times\frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}

Rationalising with a conjugate

When the denominator is $b + \sqrt{c}$ , multiply top and bottom by its conjugate $b - \sqrt{c}$ . The denominator becomes a difference of two squares, which removes the surd:

(b + \sqrt{c})(b - \sqrt{c}) = b^2 - c

Worked example

Rationalise the denominator of $\dfrac{4}{3 + \sqrt{5}}$ and simplify.

Step 1: Identify the conjugate

The denominator is $3 + \sqrt{5}$ , so its conjugate is $3 - \sqrt{5}$ .

Step 2: Multiply top and bottom by the conjugate

\frac{4}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}

Step 3: Expand the denominator as a difference of two squares

(3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4

Step 4: Simplify

\frac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5}

The surd has been removed from the denominator and the fraction cancels neatly.

Examples in context

Example 1. Exact lengths in geometry. The diagonal of a square of side $5$ is $\sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$ by Pythagoras. Leaving the answer as $5\sqrt{2}$ is exact, whereas $7.07$ is only an approximation, which is why surds appear throughout coordinate geometry and trigonometry.

Example 2. Exact trigonometric values. Standard angles give surd values, for example $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ . Manipulating such values, including rationalising fractions that contain them, is a routine surd skill that feeds straight into the trigonometry strand.

Try this

Q1. Simplify $\sqrt{45}$ . [1 mark]

Cue. $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ .

Q2. Simplify $\sqrt{8} + \sqrt{32}$ . [2 marks]

Cue. $2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}$ .

Q3. Rationalise $\dfrac{1}{2 - \sqrt{3}}$ . [3 marks]

Cue. Multiply by $\dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}$ ; denominator $= 4 - 3 = 1$ , so the answer is $2 + \sqrt{3}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSimplify

\sqrt{50} + \sqrt{18}

, giving your answer in the form

k\sqrt{2}

Show worked answer →

Write each surd with the largest square factor: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ and $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ .

Add the like surds: $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$ .

What markers reward: extracting the largest perfect-square factor from each surd, recognising both reduce to multiples of $\sqrt{2}$ , and adding the coefficients to give $8\sqrt{2}$ .

Original3 marksRationalise the denominator of

\dfrac{6}{\sqrt{3}}

and simplify.