Year 1 Pure · Chapter 1

Indices and Surds

Specification: Edexcel Pure 2 — Algebra and functions · AQA section B

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In this chapter

1. The laws of indices
2. Zero, negative and fractional indices
3. Surds
4. Rationalising the denominator
Practice

Key facts — this chapter

$a^m \times a^n = a^{m+n}$ , $\;\dfrac{a^m}{a^n} = a^{m-n}$ , $\;(a^m)^n = a^{mn}$ .
$a^0 = 1$ (for $a \neq 0$ ), $\;a^{-n} = \dfrac{1}{a^n}$ , $\;a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$ — root first, then power.
$\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ — simplify surds by pulling out the largest square factor.
To rationalise $\dfrac{k}{\sqrt{b}}$ , multiply top and bottom by $\sqrt{b}$ .
To rationalise $\dfrac{k}{a + \sqrt{b}}$ , multiply top and bottom by the conjugate $a - \sqrt{b}$ , because $(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$ .

Indices and surds are the grammar of A-level algebra — every later chapter assumes you can move between powers, roots and fractions without breaking stride. The good news: there are only a handful of rules, and every one of them exists for a reason you can see.

1. The laws of indices

A power is just a count of repeated factors: $2^3$ means $2 \times 2 \times 2$ . Keep that picture in mind and the laws stop being things to memorise — they become things you could reinvent on the spot.

The index laws

For any base $a$ and indices $m$ , $n$ :

a^m \times a^n = a^{m+n} \qquad\quad \frac{a^m}{a^n} = a^{m-n} \qquad\quad (a^m)^n = a^{mn}

The first two need the same base; you can’t combine $2^3 \times 5^2$ this way.

Why it works

Multiplying $a^3 \times a^4$ just lines up $3$ copies of $a$ next to $4$ more — seven copies in total, so $a^{3+4}$ . Dividing cancels copies pair by pair, which subtracts the counts. And $(a^3)^4$ is four groups of three copies: $3 \times 4$ copies altogether.

Why do the index laws work? Pick a base and two powers, and watch the factors do the bookkeeping — including what happens when division runs out of factors.

base operation powers

Worked example 1.1 · Simplifying with the index laws

Simplify $\left(2x^2\right)^3 \times 3x^{-4}$ .

Cube the whole bracket — the $2$ as well as the $x^2$ :
$\left(2x^2\right)^3 = 2^3 \times x^{2 \times 3} = 8x^6$
Multiply the coefficients and add the indices:
$8x^6 \times 3x^{-4} = 24x^{6+(-4)} = 24x^2$

Common mistake

$(2x^2)^3$ is $8x^6$ , not $2x^6$ — the power applies to everything inside the bracket, coefficient included. It’s the single most common indices slip at A-level.

2. Zero, negative and fractional indices

What should $a^0$ or $a^{\frac{1}{2}}$ mean? You can’t repeat a factor “half a time” — instead, we choose the meanings that keep the index laws working, and those choices turn out to be forced.

Key result

a^0 = 1 \;\;(a \neq 0) \qquad\quad a^{-n} = \frac{1}{a^n} \qquad\quad a^{\frac{1}{n}} = \sqrt[n]{a} \qquad\quad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m

Why it works

The laws leave no other option. Since $a^n \times a^{-n} = a^{n-n} = a^0$ , and dividing anything by itself gives $1$ , we need $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$ . Since $\left(a^{\frac{1}{2}}\right)^2 = a^{\frac{1}{2} \times 2} = a$ , the number $a^{\frac{1}{2}}$ squares to $a$ — it is $\sqrt{a}$ .

Worked example 1.2 · Evaluating negative and fractional indices

Evaluate $\;8^{\frac{2}{3}}$ , $\;25^{-\frac{1}{2}}$ and $\left(\frac{16}{9}\right)^{-\frac{1}{2}}$ .

For $8^{\frac{2}{3}}$ , take the root first, then the power:
$8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$
For $25^{-\frac{1}{2}}$ , the minus sign means “one over”:
$25^{-\frac{1}{2}} = \frac{1}{25^{\frac{1}{2}}} = \frac{1}{5}$
For a fraction, a negative power flips it first:
$\left(\tfrac{16}{9}\right)^{-\frac{1}{2}} = \left(\tfrac{9}{16}\right)^{\frac{1}{2}} = \frac{3}{4}$

Common mistake

Root first, then power — $\left(\sqrt[3]{8}\right)^2 = 4$ is far friendlier than $\sqrt[3]{64}$ , even though both work. And a negative index makes a reciprocal, never a negative number: $25^{-\frac{1}{2}}$ is $\frac{1}{5}$ , not $-5$ .

3. Surds

A surd is a root that doesn’t simplify to a whole number or fraction — $\sqrt{2}$ , $\sqrt{3}$ , $5\sqrt{2}$ . We leave them as they are because they’re exact: writing $1.414$ throws information away, and exam mark schemes notice.

Key result

\sqrt{ab} = \sqrt{a} \times \sqrt{b} \qquad\quad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

To simplify a surd, hunt for the largest square factor hiding inside it.

Worked example 1.3 · Simplifying surds

Simplify $\sqrt{75} + \sqrt{27}$ .

Pull the largest square factor out of each:
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \qquad\quad \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$
Both are now multiples of the same surd, so they add like like-terms:
$5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$

Common mistake

The rules split products, never sums: $\sqrt{9+16} = \sqrt{25} = 5$ , but $\sqrt{9} + \sqrt{16} = 7$ . There is no rule for $\sqrt{a+b}$ .

4. Rationalising the denominator

Answers are easier to compare, combine and mark when no surd sits underneath a fraction — so we rationalise: rebuild the fraction with a whole-number denominator without changing its value.

Case 1 — a single surd underneath

Multiply top and bottom by that surd:

\frac{k}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{k\sqrt{b}}{b}

Multiplying by $\frac{\sqrt{b}}{\sqrt{b}} = 1$ leaves the value untouched — it only changes how the fraction is written.

Worked example 1.4 · Rationalising a single surd

Rationalise $\dfrac{14}{\sqrt{7}}$ .

Multiply top and bottom by $\sqrt{7}$ :
$\frac{14}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{14\sqrt{7}}{7}$
Simplify the fraction: $\dfrac{14\sqrt{7}}{7} = 2\sqrt{7}$ .

Case 2 — two terms underneath: use the conjugate

For a denominator like $a + \sqrt{b}$ , multiply top and bottom by its conjugate $a - \sqrt{b}$ (same terms, opposite sign on the surd):

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

Why it works

It’s the difference of two squares. Expanding gives $a^2 - a\sqrt{b} + a\sqrt{b} - b$ : the two surd terms are equal and opposite, so they cancel each other exactly, leaving the rational number $a^2 - b$ . The conjugate is designed to make that cancellation happen.

Worked example 1.5 · Rationalising with a conjugate

Rationalise $\dfrac{6}{\sqrt{7} + 1}$ .

The conjugate of $\sqrt{7} + 1$ is $\sqrt{7} - 1$ . Multiply top and bottom by it:
$\frac{6}{\sqrt{7} + 1} \times \frac{\sqrt{7} - 1}{\sqrt{7} - 1} = \frac{6(\sqrt{7} - 1)}{(\sqrt{7})^2 - 1^2}$
The denominator collapses to a whole number:
$(\sqrt{7})^2 - 1^2 = 7 - 1 = 6$
Cancel the $6$ :
$\frac{6(\sqrt{7} - 1)}{6} = \sqrt{7} - 1$

Worked example 1.6 · A harder conjugate

Write $\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}$ in the form $a + b\sqrt{3}$ , where $a$ and $b$ are integers.

Multiply top and bottom by the conjugate $2 + \sqrt{3}$ :
$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 + \sqrt{3})^2}{2^2 - 3}$
Expand the top and simplify the bottom:
$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \qquad\quad 2^2 - 3 = 1$
Dividing by $1$ changes nothing:
$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = 7 + 4\sqrt{3}$

So $a = 7$ and $b = 4$ .

Common mistake

Multiply the top and bottom — rationalising is multiplying by a clever form of $1$ , and forgetting the numerator changes the value. And the conjugate flips only the sign of the surd term: the conjugate of $4 - \sqrt{2}$ is $4 + \sqrt{2}$ , not $-4 + \sqrt{2}$ .

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Write each of the following as a single power:

$2^3 \times 2^4$
$\dfrac{3^5}{3^2}$
$\left(5^2\right)^3$

Show solution

(a) Same base, multiplying — add the indices: $2^3 \times 2^4 = 2^{3+4} = 2^7$ .

(b) Same base, dividing — subtract: $\dfrac{3^5}{3^2} = 3^{5-2} = 3^3$ .

(c) Power of a power — multiply: $\left(5^2\right)^3 = 5^{2 \times 3} = 5^6$ .

2 [3 marks]

Evaluate:

$7^0$
$4^{-2}$
$27^{\frac{1}{3}}$

Show solution

(a) Anything (non-zero) to the power $0$ is $1$ : $\;7^0 = 1$ .

(b) Negative index means reciprocal: $\;4^{-2} = \dfrac{1}{4^2} = \dfrac{1}{16}$ .

(c) A $\frac{1}{3}$ power is a cube root: $\;27^{\frac{1}{3}} = \sqrt[3]{27} = 3$ .

3 [3 marks]

Simplify:

$\sqrt{48}$
$\sqrt{18} + \sqrt{8}$

Show solution

(a) The largest square factor of $48$ is $16$ :

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

(b) Simplify each surd, then add like terms:

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

4 [2 marks]

Simplify:

$\sqrt{3} \times \sqrt{12}$
$\dfrac{\sqrt{40}}{\sqrt{10}}$

Show solution

(a) Multiply under one root: $\sqrt{3} \times \sqrt{12} = \sqrt{36} = 6$ .

(b) Divide under one root: $\dfrac{\sqrt{40}}{\sqrt{10}} = \sqrt{4} = 2$ .

Standard

5 [3 marks]

Simplify $\left(2x^2y^3\right)^2 \times 3x^{-4}y$ , giving your answer without negative indices.

Show solution

Square the whole bracket — coefficient included:
$\left(2x^2y^3\right)^2 = 4x^4y^6$
Multiply coefficients and add the indices of each letter:
$4x^4y^6 \times 3x^{-4}y = 12\,x^{4 + (-4)}\,y^{6+1} = 12x^0y^7 = 12y^7$

6 [4 marks]

Evaluate:

$8^{\frac{2}{3}}$
$32^{-\frac{3}{5}}$
$\left(\dfrac{9}{4}\right)^{-\frac{1}{2}}$

Show solution

(a) Root first, then power: $8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$ .

(b) Root, power, then reciprocal:

32^{-\frac{3}{5}} = \frac{1}{\left(\sqrt[5]{32}\right)^3} = \frac{1}{2^3} = \frac{1}{8}

(c) The negative power flips the fraction first:

\left(\tfrac{9}{4}\right)^{-\frac{1}{2}} = \left(\tfrac{4}{9}\right)^{\frac{1}{2}} = \frac{2}{3}

7 [3 marks]

Expand and simplify $\left(5 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)$ .

Show solution

Expand all four terms:

5 \times 2 + 5\sqrt{3} - 2\sqrt{3} - \left(\sqrt{3}\right)^2 = 10 + 5\sqrt{3} - 2\sqrt{3} - 3

Collect the rational and surd parts separately:

(10 - 3) + (5 - 2)\sqrt{3} = 7 + 3\sqrt{3}

8 [3 marks]

Rationalise the denominator of each fraction, simplifying fully:

$\dfrac{6}{\sqrt{3}}$
$\dfrac{10}{\sqrt{5}}$

Show solution

(a) Multiply top and bottom by $\sqrt{3}$ :

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

(b) Multiply top and bottom by $\sqrt{5}$ :

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

9 [3 marks]

Solve $2^{x+1} = 8^{x-1}$ .

Show solution

Write both sides with the same base — $8 = 2^3$ :
$2^{x+1} = \left(2^3\right)^{x-1} = 2^{3x-3}$
Same base on both sides, so the indices must be equal:
$x + 1 = 3x - 3$
Solve: $4 = 2x$ , so $x = 2$ .

(Check: $2^3 = 8$ and $8^1 = 8$ . ✓)

Exam style

10 [5 marks]

Rationalise the denominator of each fraction, giving each answer in its simplest form:

$\dfrac{5}{\sqrt{7} - 2}$ [2]
$\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}$ [3]

Show solution

(a) Multiply top and bottom by the conjugate $\sqrt{7} + 2$ :

\frac{5}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{5\left(\sqrt{7} + 2\right)}{7 - 4} = \frac{5\sqrt{7} + 10}{3}

(b) Multiply top and bottom by the conjugate $3 + \sqrt{2}$ :

\frac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{\left(3 + \sqrt{2}\right)^2}{9 - 2}

Expanding the top: $\left(3 + \sqrt{2}\right)^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}$ , so the answer is

\frac{11 + 6\sqrt{2}}{7}

11 [5 marks]

Write $\left(2 + \sqrt{5}\right)^2$ in the form $a + b\sqrt{5}$ , where $a$ and $b$ are integers. [2]
Hence, or otherwise, show that $\dfrac{1}{\left(2 + \sqrt{5}\right)^2} = 9 - 4\sqrt{5}$ . [3]

Show solution

(a) Expand:

\left(2 + \sqrt{5}\right)^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}

(b) Using part (a), rationalise with the conjugate $9 - 4\sqrt{5}$ :

\frac{1}{9 + 4\sqrt{5}} \times \frac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} = \frac{9 - 4\sqrt{5}}{81 - 80} = 9 - 4\sqrt{5} \;\; \blacksquare

(The denominator $81 - 16 \times 5 = 1$ is what makes this one tidy.)

12 [4 marks]

A rectangle has width $\left(5 - \sqrt{5}\right)$ cm and area $10$ cm $^2$ . Find its length in the form $a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers.

Show solution

Length = area ÷ width:
$\text{length} = \frac{10}{5 - \sqrt{5}}$
Rationalise with the conjugate $5 + \sqrt{5}$ :
$\frac{10}{5 - \sqrt{5}} \times \frac{5 + \sqrt{5}}{5 + \sqrt{5}} = \frac{10\left(5 + \sqrt{5}\right)}{25 - 5} = \frac{50 + 10\sqrt{5}}{20}$
Simplify the fraction:
$\text{length} = \frac{5 + \sqrt{5}}{2} = \frac{5}{2} + \frac{1}{2}\sqrt{5} \text{ cm}$

So $a = \frac{5}{2}$ and $b = \frac{1}{2}$ .