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G7.1 Surds

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01 — Explanation

What makes a root a surd?

A surd is a root that cannot be written as a fraction — its decimal goes on forever with no pattern. $\sqrt{16} = 4$ is rational, so it isn't a surd, but $\sqrt{2}$ , $\sqrt{7}$ , and $\sqrt{21}$ are all surds. Recognising which roots simplify to whole numbers is the first skill: check whether the number under the root is a perfect square.

02 — Explanation

How do I simplify a surd?

Find the largest square factor of the number under the root, then split it.

\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16}\times\sqrt{3} = 4\sqrt{3}

Exam Tip: Always check for the largest square factor. Writing $\sqrt{48}=\sqrt{4\times12}=2\sqrt{12}$ is correct but incomplete — $\sqrt{12}$ still simplifies further, and you lose the final mark for stopping early.

A coefficient out front just multiplies through: $5\sqrt{27} = 5\times 3\sqrt{3} = 15\sqrt{3}$ .

03 — Explanation

How do I add or subtract surds?

Only like surds (the same number under the root) can combine — $\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}$ . Simplify each term first, then collect like terms exactly as you would with $x$ 's:

\sqrt{8}+\sqrt{18} = 2\sqrt{2}+3\sqrt{2} = 5\sqrt{2}

04 — Explanation

How do I expand brackets containing surds?

Treat $\sqrt{a}$ like any other term and use $\sqrt{a}\times\sqrt{a}=a$ to clear roots that multiply together.

(2+\sqrt5)(1-\sqrt5) = 2 - 2\sqrt5+\sqrt5-5 = -3-\sqrt5

Exam Tip: Squaring a bracket like $(3-\sqrt2)^2$ is the most common place marks are dropped. Write it as $(3-\sqrt2)(3-\sqrt2)$ and expand fully — $3^2+(\sqrt2)^2=11$ skips the middle term and is wrong. Done properly: $9-3\sqrt2-3\sqrt2+2=11-6\sqrt2$ .

05 — Explanation

What is a conjugate, and why does it matter?

$(a+\sqrt b)$ and $(a-\sqrt b)$ are conjugates. Multiplying a binomial surd by its conjugate always eliminates the root, because it's a difference of two squares:

(2+\sqrt3)(2-\sqrt3) = 2^2-(\sqrt3)^2 = 4-3=1

This is the entire mechanism behind rationalising a binomial denominator.

06 — Explanation

How do I rationalise a denominator?

Single surd — multiply top and bottom by that surd:

\frac{3}{\sqrt6}\times\frac{\sqrt6}{\sqrt6}=\frac{3\sqrt6}{6}=\frac{\sqrt6}{2}

Binomial surd — multiply top and bottom by the denominator's conjugate, then simplify:

\frac{5+2\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}

Numerator: $10-5\sqrt3+4\sqrt3-2(3) = 4-\sqrt3$ . Denominator: $4-3=1$ . Result: $4-\sqrt3$ .

Exam Tip: In a "show that" question every line must be shown — a missing step in the FOIL expansion loses marks even if your final answer is right.

07 — Worked example

Worked exam-style question

A right-angled triangle has legs $\sqrt5$ cm and $3$ cm. Find the hypotenuse in simplest surd form, then find the area of a rectangle with sides $(3+\sqrt5)$ cm and $(3-\sqrt5)$ cm.

Pythagoras: $c^2 = (\sqrt5)^2+3^2 = 5+9=14$ , so $c=\sqrt{14}$ (14 has no square factors, so this is fully simplified).
Area $=(3+\sqrt5)(3-\sqrt5) = 3^2-(\sqrt5)^2 = 9-5=4\text{ cm}^2$ — two irrational lengths producing a rational area, by the difference-of-two-squares pattern.

08 — Common mistakes

Common mistakes to avoid

Writing $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$ — this is never true.
Stopping at a partially simplified surd like $2\sqrt{12}$ instead of $4\sqrt3$ .
Sign slips when expanding a conjugate pair, e.g. treating $(1+\sqrt5)(1-\sqrt5)$ as $1-5=4$ instead of $-4$ .
Leaving a fraction like $\dfrac{6+2\sqrt3}{2}$ unsimplified — divide every term by 2 to get $3+\sqrt3$ .