SurdsEdexcel IGCSE Maths: Revision notes

A surd is a root (usually a square root) that cannot be simplified to a whole number or exact decimal — for example $\sqrt{2}$, $\sqrt{5}$, $\sqrt{7}$. Numbers like $\sqrt{9}=3$ are not surds because they simplify exactly.

Surds are irrational numbers — their decimal expansions never terminate or repeat. Exam questions ask you to give answers "in surd form" (exact) rather than as a rounded decimal, because rounding loses accuracy.

Key surd rules you must know:

• $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
• $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$
• $\sqrt{a} + \sqrt{a} = 2\sqrt{a}$ (like surds add like fractions — only combine identical surds)

To simplify a surd, find the largest square number that divides exactly into the number under the root, then split it using $\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$.

Worked method for $\sqrt{72}$:

1. Find the largest square factor of 72: that is 36 (since $72 = 36 \times 2$)
2. Split: $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2}$
3. Simplify: $= 6\sqrt{2}$

Always check your final surd cannot be simplified further — keep dividing out square factors until none remain.

Multiply surds the same way as normal numbers, using $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$, then simplify.

Example: $\sqrt{3}\times\sqrt{12} = \sqrt{36} = 6$

When expanding brackets containing surds, use the standard distributive law (and FOIL for double brackets), remembering $\sqrt{a}\times\sqrt{a}=a$.

Example: $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 2\sqrt{3} + 2\sqrt{3} - 3 = 4 - 3 = 1$

Notice this is a difference of two squares: $(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$. This pattern is the key to rationalising denominators with two-term surds.

A fraction like $\dfrac{1}{\sqrt{2}}$ has an irrational denominator. Rationalising means rewriting it with a rational (whole number) denominator, without changing its value.

Case 1 — single surd on the denominator: multiply top and bottom by that surd. $\dfrac{1}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$

Case 2 — denominator is $a+\sqrt{b}$ (a binomial): multiply top and bottom by the conjugate to trigger the difference of two squares, which eliminates the surd. $\dfrac{1}{3+\sqrt{2}} = \dfrac{1}{3+\sqrt{2}}\times\dfrac{3-\sqrt{2}}{3-\sqrt{2}} = \dfrac{3-\sqrt{2}}{9-2} = \dfrac{3-\sqrt{2}}{7}$

Always multiply by a fraction equal to 1 (same surd or conjugate top and bottom) — this guarantees the value is unchanged.

Index laws extend naturally to fractional and negative powers — these appear constantly alongside surds.

Fractional indices (roots):

• $a^{\frac{1}{n}} = \sqrt[n]{a}$ (e.g. $a^{\frac{1}{2}}=\sqrt{a}$, $a^{\frac{1}{3}}=\sqrt[3]{a}$)
• $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}$

Negative indices (reciprocals):

• $a^{-n} = \dfrac{1}{a^{n}}$

Standard index laws (still apply):

Worked example: $27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9$

Worked example: $4^{-\frac{1}{2}} = \dfrac{1}{4^{\frac{1}{2}}} = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}$