Surds & Indices

Why This Matters

Pick up your calculator and type $\sqrt{2}$. You get $1.41421356\ldots$ — a decimal that never ends and never repeats. You cannot write it down exactly as a decimal. But you can write it exactly as $\sqrt{2}$.

That is what surds are for: exact values. A-Level exams almost always want exact answers, so you need to be fluent with surds and indices from day one.

Laws of Indices

You met these at GCSE. At A-Level, they must be second nature — especially with fractional and negative exponents.

The core laws

For any base $a \neq 0$ and any real exponents $m$ and $n$:

Watch out: $a^0 = 1$ is true for every non-zero value of $a$. It is not “undefined” and it is not $0$. If you are not sure why, notice that $a^3 \div a^3 = a^{3-3} = a^0$, and anything divided by itself is $1$.

Negative indices are NOT negative numbers

This is the single most common mistake in this topic. Compare:

$x^{-2} = \frac{1}{x^2} \qquad \text{(always positive when } x \neq 0\text{)}$

$-x^2 \qquad \text{(always negative when } x \neq 0\text{)}$

These are completely different expressions. A negative exponent means “reciprocal”, not “make it negative”.

Fractional indices ARE roots

The connection is simple: the denominator of the fraction is the root, and the numerator is the power.

$x^{\frac{1}{2}} = \sqrt{x}, \qquad x^{\frac{1}{3}} = \sqrt[3]{x}, \qquad x^{\frac{3}{2}} = (\sqrt{x})^3 = \sqrt{x^3}$

Either order works — root first then power, or power first then root. Root first is usually easier with numbers.

Surds

A surd is a root that cannot be simplified to a rational number. For example, $\sqrt{2}$, $\sqrt{5}$, and $\sqrt{11}$ are surds. But $\sqrt{9} = 3$ is not a surd — it simplifies to a whole number.

Simplifying surds

The key property is:

$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

To simplify a surd, find the largest square factor inside the root.

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

$\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$

The surd trap: $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. Try it: $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$. They are not equal. The square root does not “distribute” over addition.

Multiplying and dividing surds

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \qquad \text{and} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$

These rules only work with multiplication and division — never with addition or subtraction.

Adding and subtracting surds

You can only combine like surds, just as you combine like terms in algebra:

$3\sqrt{5} + 7\sqrt{5} = 10\sqrt{5}$

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

But $3\sqrt{2} + 4\sqrt{3}$ cannot be simplified further — they are unlike surds.

Rationalising the Denominator

A-Level convention is: no surds in the denominator. Rationalising removes them.

Type 1: Single surd in the denominator

Multiply top and bottom by the surd:

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

Type 2: Two terms in the denominator

When the denominator is $a + \sqrt{b}$ or $a - \sqrt{b}$, multiply by the conjugate. The conjugate flips the sign of the surd term.

$\frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$

This works because of the difference of two squares: $(a + b)(a - b) = a^2 - b^2$. The surd disappears from the bottom.

Remember: you must multiply both the numerator and the denominator by the conjugate. Multiplying only the bottom changes the value of the fraction.

Worked Examples

Example 1: Index laws (warm-up)

Simplify $\frac{x^5 \cdot x^3}{x^2}$

Add the indices on top: $x^5 \cdot x^3 = x^8$

Subtract the index on the bottom: $\frac{x^8}{x^2} = x^{8-2} = x^6$

Example 2: Fractional and negative indices

Evaluate $8^{\frac{2}{3}}$ and write $\frac{3}{x^4}$ in the form $3x^n$

For $8^{\frac{2}{3}}$: root first, then power.

$8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$

For $\frac{3}{x^4}$: bring $x^4$ up with a negative index.

$\frac{3}{x^4} = 3x^{-4}$

Example 3: Simplifying a surd expression

Simplify $\sqrt{50} + 3\sqrt{8} - \sqrt{32}$

Break each surd into its simplest form:

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

$3\sqrt{8} = 3\sqrt{4 \cdot 2} = 3 \cdot 2\sqrt{2} = 6\sqrt{2}$

$\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$

Now combine like surds:

$5\sqrt{2} + 6\sqrt{2} - 4\sqrt{2} = 7\sqrt{2}$

Example 4: Rationalising with a conjugate

Express $\frac{4}{3 - \sqrt{5}}$ in the form $a + b\sqrt{5}$

Multiply top and bottom by the conjugate $3 + \sqrt{5}$:

$\frac{4}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{4(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2}$

$= \frac{12 + 4\sqrt{5}}{9 - 5} = \frac{12 + 4\sqrt{5}}{4} = 3 + \sqrt{5}$

So $a = 3$ and $b = 1$.

Example 5: Exam-style mixed question

Given that $\frac{2\sqrt{3} + \sqrt{6}}{\sqrt{3}} = a + \sqrt{b}$, find the values of $a$ and $b$.

Split the fraction into two parts:

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

So $a = 2$ and $b = 2$.

Alternatively, you could rationalise by multiplying top and bottom by $\sqrt{3}$:

$\frac{(2\sqrt{3} + \sqrt{6}) \cdot \sqrt{3}}{3} = \frac{2 \cdot 3 + \sqrt{18}}{3} = \frac{6 + 3\sqrt{2}}{3} = 2 + \sqrt{2}$

Both methods give the same answer. Choose whichever feels cleaner.

Common Patterns