Simplify surds, perform the four operations on surds, and rationalise denominators including those of the form a plus root b

What this dot point is asking

SEAB wants you to treat surds (irrational square roots such as $\sqrt{2}$ or $\sqrt{3}$) as exact numbers, to simplify them, to add, subtract, multiply and divide them, and to rationalise a denominator so that no surd is left underneath. This matters because exact answers in surd form are demanded throughout A-Maths, and a tidy surd is often the cleaner answer the examiner wants.

The answer

What a surd is

A surd is a root that cannot be written as an exact fraction, so $\sqrt{4} = 2$ is not a surd, but $\sqrt{2}$ is. We keep surds exact rather than rounding to a decimal. The two rules that drive every simplification are:

Simplifying a surd

To simplify, pull out the largest perfect-square factor:

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

Adding and subtracting surds

You may only combine like surds, the way you collect like terms in algebra. So $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$, but $3\sqrt{2} + 5\sqrt{3}$ cannot be combined. Always simplify each surd first, because unlike surds often become like once simplified.

Rationalising the denominator

A fraction should not be left with a surd on the bottom. If the denominator is a single surd, multiply top and bottom by that surd:

If the denominator has the form $a + \sqrt{b}$, multiply by its conjugate $a - \sqrt{b}$. The product of conjugates is a difference of two squares, which removes the surd:

Examples in context

Example 1. Exact lengths in geometry. The diagonal of a unit square is $\sqrt{2}$, and the height of an equilateral triangle of side $2$ is $\sqrt{3}$. Leaving these as surds keeps later area and Pythagoras calculations exact, which is why coordinate-geometry answers so often appear in surd form.

Example 2. Rationalising before differentiating. A function such as $y = \dfrac{1}{\sqrt{x}}$ is easier to differentiate once rewritten as $y = x^{-1/2}$, the same idea of clearing a root from the denominator that you meet in surd manipulation.

Try this

Q1. Simplify $\sqrt{75} - \sqrt{12}$. [2 marks]

• Cue. $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{12} = 2\sqrt{3}$, so the answer is $3\sqrt{3}$.

Q2. Rationalise $\dfrac{5}{\sqrt{5}}$. [2 marks]

• Cue. Multiply top and bottom by $\sqrt{5}$: $\dfrac{5\sqrt{5}}{5} = \sqrt{5}$.

Q3. Express $\dfrac{1}{2 + \sqrt{3}}$ in the form $a + b\sqrt{3}$. [3 marks]

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