Surd Manipulation and Rationalising Denominators Revision Notes | GCSE AQA Higher Further Maths

1. Why Surds Matter: The Case for Exact Answers

Calculators give you $\sqrt{2} \approx 1.41421...$, but that decimal never ends and never repeats. The moment you round it, your answer is wrong — slightly, but wrong. Surds let you carry the exact value all the way through a calculation without any rounding error creeping in.

This matters most in geometry and trigonometry, where one rounded intermediate value can throw off a final answer by a visible amount. Exam questions that say "give your answer in exact form" or "show that" are specifically testing whether you can work with surds rather than reaching for the decimal button.

2. The Core Rules for Manipulating Surds

Two multiplication rules underpin almost everything you do with surds. They follow directly from index laws, so if you are comfortable with those, these will feel familiar.

The most useful application of the product rule is simplifying surds: rewrite $\sqrt{n}$ by splitting $n$ into a perfect-square factor and whatever is left. For example, $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$. You always want the largest perfect-square factor so you reach the simplest form in one step.

Worked example: Simplifying a Surd

Simplify $\sqrt{72}$. (Non-calculator)

1. $72 = 36 \times 2$ — 36 is the largest perfect square that divides 72. Using 4 × 18 or 9 × 8 would also work but takes more steps. [M1 for correct factor pair]
2. $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2}$ — Split the root across the two factors using $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
3. $\sqrt{36} = 6$ — $6 \times 6 = 36$, so $\sqrt{36} = 6$. [A1 for correct simplified form]
4. $\sqrt{72} = 6\sqrt{2}$ — Check: $(6\sqrt{2})^2 = 36 \times 2 = 72$. Correct.

Answer: $6\sqrt{2}$

3. Adding, Subtracting, and Expanding with Surds

Surds add and subtract like algebraic terms: you can only combine like surds. Think of $\sqrt{3}$ as a unit — $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$ for the same reason that $2x + 5x = 7x$. You cannot add $\sqrt{2}$ and $\sqrt{3}$ any more than you can add $x$ and $y$.

When expanding brackets, you use the same distributive law as in algebra. The only extra step is applying $(\sqrt{a})^2 = a$ whenever a surd is squared during the expansion.

Worked example: Expanding Brackets Containing Surds

Expand and simplify $(3 + \sqrt{5})(2 - \sqrt{5})$. (Non-calculator)

1. $(3)(2) + (3)(-\sqrt{5}) + (\sqrt{5})(2) + (\sqrt{5})(-\sqrt{5})$ — Multiply each term in the first bracket by each term in the second. [M1 for correct expansion attempt]
2. $= 6 - 3\sqrt{5} + 2\sqrt{5} - (\sqrt{5})^2$ — Collect the four separate products before simplifying.
3. $= 6 - 3\sqrt{5} + 2\sqrt{5} - 5$ — A surd squared equals the number under the root sign. This is the step most students miss.
4. $= (6 - 5) + (-3\sqrt{5} + 2\sqrt{5}) = 1 - \sqrt{5}$ — Rational terms: $6 - 5 = 1$. Surd terms: $-3\sqrt{5} + 2\sqrt{5} = -\sqrt{5}$. [A1 for correct final answer]

Answer: $1 - \sqrt{5}$

4. Rationalising the Denominator

A fraction with a surd in the denominator is not in its simplest form. Rationalising the denominator means rewriting the fraction so the denominator is a rational number — no surds allowed at the bottom. This is a required step before leaving a final answer in most GCSE questions.

There are two cases to handle: a denominator with a single surd term, and a denominator with two terms (one of which involves a surd). Each case has its own technique.

4a. Single-Term Denominator: $\frac{a}{\sqrt{b}}$

Multiply the top and bottom of the fraction by $\sqrt{b}$. This is equivalent to multiplying by 1, so the value of the fraction does not change. The denominator becomes $(\sqrt{b})^2 = b$, which is rational.

Worked example: Rationalising a Single-Term Denominator

Rationalise the denominator of $\dfrac{6}{\sqrt{3}}$. Give your answer in simplified form. (Non-calculator)

1. $\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$ — Multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$ is multiplying by 1 — the value is unchanged. [M1 for correct multiplier]
2. $= \dfrac{6\sqrt{3}}{\sqrt{3} \times \sqrt{3}}$ — Numerator: $6 \times \sqrt{3} = 6\sqrt{3}$.
3. $\sqrt{3} \times \sqrt{3} = (\sqrt{3})^2 = 3$ — The surd disappears from the denominator. This is the whole point of the technique.
4. $= \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$ — $6 \div 3 = 2$. [A1 for correct simplified answer]

Answer: $2\sqrt{3}$

4b. Two-Term Denominator: $\frac{a}{p + q\sqrt{b}}$

When the denominator has two terms, multiplying by $\sqrt{b}$ alone will not clear the surd — you will still have a surd term left. Instead, multiply by the conjugate of the denominator. The conjugate of $(p + q\sqrt{b})$ is $(p - q\sqrt{b})$: you flip the sign between the two terms.

The product of a pair of conjugates uses the difference of two squares: $(p + q\sqrt{b})(p - q\sqrt{b}) = p^2 - q^2 b$. Both terms are rational, so the surd vanishes completely from the denominator.

Worked example: Rationalising a Two-Term Denominator

Rationalise the denominator of $\dfrac{4}{3 + \sqrt{2}}$. Give your answer in simplified form. (Non-calculator)

1. Denominator: $3 + \sqrt{2}$, so conjugate is $3 - \sqrt{2}$ — Flip the sign between the two terms. [M1 for using the correct conjugate]
2. $\dfrac{4}{3 + \sqrt{2}} \times \dfrac{3 - \sqrt{2}}{3 - \sqrt{2}}$ — Again, multiplying by $\frac{3-\sqrt{2}}{3-\sqrt{2}}$ equals multiplying by 1.
3. $(3 + \sqrt{2})(3 - \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7$ — $3^2 = 9$ and $(\sqrt{2})^2 = 2$. The surd terms cancel. [M1 for correct denominator]
4. $4(3 - \sqrt{2}) = 12 - 4\sqrt{2}$ — Distribute the 4 across both terms in the conjugate.
5. $\dfrac{12 - 4\sqrt{2}}{7}$ — Check whether the fraction simplifies further: 12, 4, and 7 share no common factor, so this is fully simplified. [A1 for correct final answer]

Answer: $\dfrac{12 - 4\sqrt{2}}{7}$

5. Surds in Exact Calculations

Surds appear naturally whenever you apply Pythagoras' theorem or trigonometry to triangles that do not have nice integer sides. Keeping the surd form throughout the working — rather than converting to a decimal — preserves accuracy and is often required by the question.

Worked example: Exact Calculation Using Surds

A right-angled triangle has legs of length $\sqrt{5}$ cm and $\sqrt{7}$ cm. Find the exact length of the hypotenuse. (Non-calculator)

1. $h^2 = (\sqrt{5})^2 + (\sqrt{7})^2$ — For a right-angled triangle: hypotenuse squared equals the sum of the squares of the other two sides. [M1]
2. $(\sqrt{5})^2 = 5$ and $(\sqrt{7})^2 = 7$ — A surd squared gives the number under the root. No calculator needed.
3. $h^2 = 5 + 7 = 12$ — The calculation stays entirely in rational numbers at this stage.
4. $h = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ — Split $\sqrt{12}$ using the largest perfect-square factor: $\sqrt{4} = 2$. [A1 for $2\sqrt{3}$ in simplified form]

Answer: $2\sqrt{3}$ cm

Notice how the answer $2\sqrt{3}$ cm is exact, whereas $\sqrt{12} \approx 3.464$ cm is a rounded approximation. If the question had asked for the area of a square with this hypotenuse as its side, you would square $2\sqrt{3}$ to get exactly $12$ cm² — no rounding error at all.

6. Exam Technique and Common Mistakes

Surd questions reward careful, methodical working. The marks are split between method and accuracy, so even if you make an arithmetic slip, clear working protects your method marks. Here is where students most often drop marks — and how to avoid each trap.

Surd manipulation and rationalising the denominator together give you the tools to work with exact values throughout any calculation — the two core skills the specification requires. Practise spotting perfect-square factors quickly and writing conjugates without hesitation, and these questions become reliable mark-winners on Paper 1.