Simplifying and working with surds (simplifying, adding and subtracting, expanding brackets, rationalising the denominator) and applying the laws of indices including negative and fractional indices.

What this dot point is asking

The SQA wants you to simplify a surd, add and subtract like surds, expand brackets containing surds, rationalise a denominator, and apply the laws of indices including negative and fractional powers. All of this must be done exactly, by hand, because surds and indices are a Paper 1 (non-calculator) staple.

Simplifying surds

A surd is a root that cannot be written exactly as a whole number or fraction, such as $\sqrt{2}$ or $\sqrt{5}$. To simplify, find the largest perfect-square factor of the number under the root and split the surd using $\sqrt{ab} = \sqrt{a}\sqrt{b}$.

Adding and subtracting surds

You can only add or subtract surds that have the same number under the root (like surds), exactly as you collect like terms in algebra. Often you must simplify each surd first so the like terms appear.

Expanding brackets with surds

Treat the surd like an algebraic term and multiply out, using $\sqrt{a} \times \sqrt{a} = a$.

Rationalising the denominator

A fraction is not in its simplest exact form while a surd sits on the bottom. To rationalise $\dfrac{a}{\sqrt{b}}$, multiply the top and bottom by $\sqrt{b}$, because $\sqrt{b} \times \sqrt{b} = b$ removes the surd from the denominator.

For example, $\dfrac{10}{\sqrt{5}} = \dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$ after simplifying the resulting fraction.

The laws of indices

The same rules that govern numbers like $2^3$ govern algebraic powers. Knowing them lets you simplify expressions and switch between root and power notation.

A negative index means "one over": $5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}$. A fractional index is a root: the denominator is the root and the numerator is the power, so $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$.

Examples in context

Surd answers appear whenever an exact length is needed. The diagonal of a square of side $4$ cm is $\sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$ cm by Pythagoras, an exact value that no calculator rounding can improve on. Fractional indices model growth and scaling: a quantity that doubles over a fixed period is multiplied by $2^{t}$, and a value such as $2^{1/2} = \sqrt{2}$ describes the multiplier over half a period.

Try this

Q1. Simplify $\sqrt{45} + \sqrt{20}$. [2 marks]

• Cue. $3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}$.

Q2. Rationalise the denominator of $\dfrac{8}{\sqrt{2}}$. [2 marks]

• Cue. $\dfrac{8\sqrt{2}}{2} = 4\sqrt{2}$.

Q3. Evaluate $27^{2/3}$. [2 marks]

• Cue. $(\sqrt[3]{27})^2 = 3^2 = 9$.