Surds

A surd is an irrational root — a root that cannot be expressed as a rational number (a fraction of two integers).

$\sqrt{4}=2$ — rational, not a surd.

$\sqrt{2}=1.41421\dots$ — irrational, a surd.

$\sqrt[3]{8}=2$ — rational, not a surd.

$\sqrt[3]{5}=1.70997\dots$ — irrational, a surd.

Recognising surds at GCSE: $\sqrt{n}$ is a surd whenever $n$ is a positive integer that is not a perfect square. The GCSE specification focuses on square roots.

Why surds matter: using exact surd form avoids rounding errors in multi-step calculations. Answers to Edexcel Higher questions may require surd form for full marks.

Key rule: $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ and $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$ (for $a,b\ge 0$).

A surd $\sqrt{n}$ is in simplified form when $n$ has no perfect square factors (other than 1).

Method: find the largest perfect square factor of $n$, split, and evaluate the rational part.

Worked examples:

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

$\sqrt{50}=\sqrt{25\times 2}=\sqrt{25}\times \sqrt{2}=5\sqrt{2}$

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

$\sqrt{200}=\sqrt{100\times 2}=\sqrt{100}\times \sqrt{2}=10\sqrt{2}$

Choosing the largest perfect square factor in one step is more efficient. $\sqrt{72}=\sqrt{4\times 18}=2\sqrt{18}=2\sqrt{9\times 2}=6\sqrt{2}$ — this works but takes two steps.

Surds can only be added or subtracted when they have the same irrational part — like collecting like terms in algebra.

$3\sqrt{2}+5\sqrt{2}=8\sqrt{2}\text{(same surd: }\sqrt{2}\text{)}$

$7\sqrt{3}-2\sqrt{3}=5\sqrt{3}$

$\sqrt{2}+\sqrt{3}\text{— cannot be simplified (different surds)}$

Worked example — simplify $\sqrt{50}+\sqrt{18}-\sqrt{8}$.

Simplify each surd first: $\sqrt{50}=5\sqrt{2},\sqrt{18}=3\sqrt{2},\sqrt{8}=2\sqrt{2}$

Collect: $5\sqrt{2}+3\sqrt{2}-2\sqrt{2}=6\sqrt{2}$ ✓

Expand brackets involving surds exactly as in algebra — multiply every term by every other term.

Single bracket:

$\sqrt{3}(2+\sqrt{3})=2\sqrt{3}+\sqrt{3}\times \sqrt{3}=2\sqrt{3}+3$

Double brackets (FOIL):

$(1+\sqrt{5})(3-\sqrt{5})$

$=3-\sqrt{5}+3\sqrt{5}-\sqrt{5}\times \sqrt{5}$

$=3+2\sqrt{5}-5$

$=-2+2\sqrt{5}$

The difference of two squares — a key pattern:

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

This always produces a rational result — the surds cancel. It is the basis of rationalising the denominator.

Worked example: $(3+\sqrt{7})(3-\sqrt{7})=9-7=2$ ✓

A fraction with a surd in the denominator can always be rewritten with a rational denominator — this is called rationalising.

Simple denominator of the form $\sqrt{a}$: multiply numerator and denominator by $\sqrt{a}$.

$\frac{3}{\sqrt{5}}=\frac{3\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}=\frac{3\sqrt{5}}{5}$

$\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$

$\frac{10}{\sqrt{2}}=\frac{10\sqrt{2}}{2}=5\sqrt{2}$

Worked example — rationalise $\frac{4}{3\sqrt{2}}$:

$\frac{4}{3\sqrt{2}}=\frac{4\sqrt{2}}{3\times 2}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$ ✓

For denominators of the form $a+\sqrt{b}$ or $a-\sqrt{b}$, multiply by the conjugate (change the sign in the denominator).

The conjugate of $(a+\sqrt{b})$ is $(a-\sqrt{b})$, and their product is $a^2-b$ (rational).

Worked example — rationalise $\frac{5}{3+\sqrt{2}}$:

$\frac{5}{3+\sqrt{2}}\times \frac{3-\sqrt{2}}{3-\sqrt{2}}=\frac{5(3-\sqrt{2})}{9-2}=\frac{15-5\sqrt{2}}{7}$

Worked example — rationalise $\frac{4+\sqrt{3}}{2-\sqrt{3}}$:

$\frac{(4+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{8+4\sqrt{3}+2\sqrt{3}+3}{4-3}=\frac{11+6\sqrt{3}}{1}=11+6\sqrt{3}$ ✓

1. $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$

$\sqrt{9+16}=\sqrt{25}=5$, not $\sqrt{9}+\sqrt{16}=3+4=7$. The square root distributes over multiplication and division, but not over addition or subtraction.

2. Not simplifying surds before adding

$\sqrt{50}+\sqrt{18}$ cannot be added directly as $\sqrt{68}$. Simplify first: $5\sqrt{2}+3\sqrt{2}=8\sqrt{2}$.

3. Rationalising — using the same sign, not the conjugate

To rationalise $\frac{1}{3+\sqrt{2}}$, multiply by $\frac{3-\sqrt{2}}{3-\sqrt{2}}$, not $\frac{3+\sqrt{2}}{3+\sqrt{2}}$. Using the same sign gives a more complex denominator.

4. Forgetting to simplify the coefficient after rationalising

After rationalising $\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}$, simplify to $2\sqrt{3}$ — leaving $\frac{6\sqrt{3}}{3}$ is not fully simplified.

Mistake	Correction
"$\sqrt{12}=2\sqrt{6}$"	$\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$ (largest square factor is 4, not 2)
"$\sqrt{50}+\sqrt{18}=\sqrt{68}$"	Simplify first: $5\sqrt{2}+3\sqrt{2}=8\sqrt{2}$
"$(2+\sqrt{3}{)}^2=4+3$"	$(2+\sqrt{3}{)}^2=4+4\sqrt{3}+3=7+4\sqrt{3}$ (the middle term $2\times 2\times \sqrt{3}$ is missing)