Surds & indices — notes, question types & practice · Simplification

01

Overview

A power question always has the same escape route: write every number as a power of one prime. $1024$ is $2^{10}$ and also $4^5$ ; $81$ is $3^4$ and also $27^{4/3}$ , because $27^4 = 81^3$ . Once the bases match, only the exponents remain.

02

The laws that move exponents

Law	Statement
same base, product	$a^m \times a^n = a^{m+n}$
same base, quotient	$a^m \div a^n = a^{m-n}$
power of a power	$(a^m)^n = a^{mn}$
zero and negative	$a^0 = 1$ , $a^{-n} = \dfrac{1}{a^n}$
fractional index	$a^{p/q} = \sqrt[q]{a^p}$

Rule: Make every base the same prime first. Then add, subtract and multiply exponents; never touch the numbers again.

03

Solving a power equation

$4^{x+1} \times 16^{2-x} = 64$ becomes $2^{2x+2} \times 2^{8-4x} = 2^6$ , so $2x + 2 + 8 - 4x = 6$ , giving $x = 2$ . Fractional indices evaluate in the same spirit: $32^{2/5} = (2^5)^{2/5} = 2^2 = 4$ , and $81^{3/4} = 27$ .

04

Rationalising with the conjugate

Multiply top and bottom by the conjugate, the same terms with the flipped middle sign:

\dfrac{1}{\sqrt{a} \pm \sqrt{b}} = \dfrac{\sqrt{a} \mp \sqrt{b}}{a - b}

The product $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$ clears every root from the bottom.

05

Telescoping chains

A sum like $\dfrac{1}{\sqrt{9}+\sqrt{8}} + \dfrac{1}{\sqrt{8}+\sqrt{7}} + \dfrac{1}{\sqrt{7}+\sqrt{6}} + \dfrac{1}{\sqrt{6}+\sqrt{5}}$ rationalises term by term into $\sqrt{9}-\sqrt{8} + \sqrt{8}-\sqrt{7} + \dots$ , and the middle roots kill each other. Only the ends survive: $\sqrt{9} - \sqrt{5} = 3 - \sqrt{5}$ .

Tip: In a chain of $\dfrac{1}{\sqrt{k}+\sqrt{k-1}}$ terms, write the answer as first root minus last root and skip the middle entirely.

06

The a plus two root b split

To open $\sqrt{a + 2\sqrt{b}}$ , hunt two numbers with sum $a$ and product $b$ ; the root is $\sqrt{x} + \sqrt{y}$ . For $\sqrt{7 + 4\sqrt{3}}$ , write $4\sqrt{3}$ as $2\sqrt{12}$ , so $x + y = 7$ and $xy = 12$ : the pair is $4$ and $3$ , giving $2 + \sqrt{3}$ .

Watch: $\sqrt{a+b}$ is not $\sqrt{a} + \sqrt{b}$ . The split only works when the middle term is forced into the $2\sqrt{b}$ shape first.

07

Comparing surds

Raise every surd to the LCM of the root orders, then compare plain integers. For $\sqrt{3}$ , $\sqrt[3]{7}$ , $\sqrt[4]{5}$ , the orders 2, 3, 4 give LCM 12. Twelfth powers: $3^6 = 729$ , $7^3 = 343$ , $5^4 = 625$ . So $\sqrt{3}$ is the largest.

08

A surd and its reciprocal

If $x = a + \sqrt{b}$ and $a^2 - b = 1$ , then $\dfrac{1}{x} = a - \sqrt{b}$ . Check first: for $5 + 2\sqrt{6}$ , $25 - 24 = 1$ , so the reciprocal is $5 - 2\sqrt{6}$ and $x + \dfrac{1}{x} = 10$ . If $a^2 - b$ is not a perfect square, do the long rationalisation instead.

09

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Evaluate fractional and negative powers

How to spot it:

Numbers raised to fractions or negatives, like $32^{2/5}$ or $81^{3/4}$ , often mixed in one expression.

a^{p/q} = \sqrt[q]{a^p}

Method

Factor each base into primes.
Apply the fractional index: root first, then power.
Combine the results with ordinary arithmetic.

Why it works:

A fractional index is a root in disguise, and prime bases make the root visible.

Try this

Find $(32)^{2/5} + (81)^{3/4}$ .

Show solution

$32 = 2^5$ , so $32^{2/5} = 2^2 = 4$ .
$81 = 3^4$ , so $81^{3/4} = 3^3 = 27$ .
$4 + 27 = 31$ .

Answer

31

Type 2very common2 practice Q

Rationalisation and telescoping surd fractions

How to spot it:

Fractions with roots in the denominator, or a long sum of such fractions with roots stepping down by one.

\dfrac{1}{\sqrt{a}\pm\sqrt{b}} = \dfrac{\sqrt{a}\mp\sqrt{b}}{a-b}

Method

Multiply top and bottom by the conjugate.
In a long sum, rationalise every term.
Watch the middle roots cancel in pairs.
Write the answer as first root minus last root.

Why it works:

Each term becomes a difference of roots that cancels with its neighbour, leaving only the two ends.

Try this

Find $\dfrac{1}{\sqrt{9}+\sqrt{8}} + \dfrac{1}{\sqrt{8}+\sqrt{7}} + \dfrac{1}{\sqrt{7}+\sqrt{6}} + \dfrac{1}{\sqrt{6}+\sqrt{5}}$ .

Show solution

Each term equals $\sqrt{k} - \sqrt{k-1}$ .
The middle roots cancel pairwise.
Left with $\sqrt{9} - \sqrt{5} = 3 - \sqrt{5}$ .

Answer

3 - sqrt(5)

Type 3common2 practice Q

Root of a surd: the a plus two root b split

How to spot it:

A square root wrapped around a number plus a root, like $\sqrt{9 + 4\sqrt{5}}$ .

\sqrt{a + 2\sqrt{b}} = \sqrt{x} + \sqrt{y},\ x+y = a,\ xy = b

Method

Rewrite the middle term as $2\sqrt{b}$ if needed.
Find two numbers with sum $a$ and product $b$ .
Answer is $\sqrt{x} + \sqrt{y}$ , or their difference when the middle sign is minus.

Why it works:

Squaring $\sqrt{x} + \sqrt{y}$ returns exactly $x + y + 2\sqrt{xy}$ .

Try this

Simplify $\sqrt{9 + 4\sqrt{5}}$ .

Show solution

$4\sqrt{5} = 2\sqrt{20}$ , so sum $9$ , product $20$ .
The pair is $5$ and $4$ .
$\sqrt{9 + 4\sqrt{5}} = \sqrt{5} + \sqrt{4} = \sqrt{5} + 2$ .

Answer

sqrt(5) + 2

Type 4common3 practice Q

Comparing surds of different orders

How to spot it:

A list of roots with different orders, asking which is largest or smallest.

\sqrt[n]{a} > \sqrt[m]{b} \iff a^{\text{lcm}/n} > b^{\text{lcm}/m}

Method

Take the LCM of all the root orders.
Raise each surd to that LCM; roots disappear.
Compare the integers and read off the answer.

Why it works:

A common power turns every surd into an integer, and integers compare at a glance.

Try this

Which is the largest: $\sqrt{3}$ , $\sqrt[3]{7}$ or $\sqrt[4]{5}$ ?

Show solution

Orders 2, 3, 4 give LCM 12.
Twelfth powers: $3^6 = 729$ , $7^3 = 343$ , $5^4 = 625$ .
$729$ is largest, so $\sqrt{3}$ wins.

Answer

sqrt(3)

Type 5common

Solve an equation in powers

How to spot it:

An unknown sits in the exponent, and the bases are powers of one number.

a^{f(x)} = a^{g(x)} \Rightarrow f(x) = g(x)

Method

Write every base as a power of the smallest prime present.
Collect the exponents of the left side into one.
Write the right side in the same base.
Equate exponents and solve the linear equation.

Why it works:

Equal bases make the exponent equation the only content left.

Try this

If $4^{x+1} \times 16^{2-x} = 64$ , find $x$ .

Show solution

$4 = 2^2$ , $16 = 2^4$ , $64 = 2^6$ .
$2^{2x+2} \times 2^{8-4x} = 2^6$ .
$2x + 2 + 8 - 4x = 6$ , so $-2x + 10 = 6$ .
$x = 2$ .

Answer

2

10

Formula sheet

Same base

a^m \cdot a^n = a^{m+n},\quad \frac{a^m}{a^n} = a^{m-n}

Power of power

(a^m)^n = a^{mn},\quad (ab)^n = a^n b^n

Zero and negative index

a^0 = 1,\quad a^{-n} = \frac{1}{a^n}

Fractional index

a^{p/q} = \sqrt[q]{a^p}

Rationalisation

\frac{1}{\sqrt{a} \pm \sqrt{b}} = \frac{\sqrt{a} \mp \sqrt{b}}{a - b}

Root of a surd

\sqrt{a \pm 2\sqrt{b}} = \sqrt{x} \pm \sqrt{y},\ x + y = a,\ xy = b

Reciprocal of a unit surd

x = a + \sqrt{b},\ a^2 - b = 1 \Rightarrow \tfrac{1}{x} = a - \sqrt{b}

11

Shortcuts that save time

⚡ Common base, then equate powers

Write both sides as powers of the same prime. The equation becomes a linear equation in the exponent.

Example

If 3^(x+2) = 27^(x-2), find x.

Show solution

$27 = 3^3$ , so $3^{x+2} = 3^{3x-6}$ .
$x + 2 = 3x - 6$ .
$8 = 2x$ , so $x = 4$ .

Answer

4

⚡ Split a + 2 root b

Force the middle term into the 2 root b shape, then find two numbers with the given sum and product.

Example

Simplify $\sqrt{7 + 4\sqrt{3}}$ .

Show solution

$4\sqrt{3} = 2\sqrt{12}$ , so $x + y = 7$ , $xy = 12$ .
The pair is $4$ and $3$ .
$\sqrt{7 + 4\sqrt{3}} = \sqrt{4} + \sqrt{3} = 2 + \sqrt{3}$ .

Answer

2 + sqrt(3)

⚡ Reciprocal by the a squared minus b check

For x = a + sqrt(b), test a squared minus b. If it equals 1, the reciprocal is a minus sqrt(b) with no work.

Example

If x = 5 + 2*sqrt(6), find 1/x.

Show solution

$a = 5$ , $b = 24$ : $25 - 24 = 1$ .
So $\dfrac{1}{x} = 5 - 2\sqrt{6}$ .
Check: the product is $25 - 24 = 1$ .

Answer

5 - 2*sqrt(6)

12

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Writing the square root of a sum as the sum of the square roots.

Roots do not distribute over addition; only the a + 2 root b split opens a nested root.

Mistake 02

Reading a to the power minus n as minus a to the n.

A negative exponent flips the base: a to the power minus n is one over a to the n.

Mistake 03

Adding exponents when the bases differ.

Convert to one prime base first; only same-base powers add exponents.

Mistake 04

Comparing surds of different orders by their radicands.

Raise all surds to the LCM of the root orders and compare the integers.

Mistake 05

Splitting a + 2 root b without rewriting the middle term.

Force the middle term into the 2 root b shape so b is the product, not the visible number.

13

Quick revision

Read this the night before the exam.

One prime base; then exponents add, subtract or multiply.
Power of a power multiplies the exponents.
Conjugate rationalisation: flip the middle sign, divide by $a - b$ .
Telescoping chain: answer is first root minus last root.
$\sqrt{a + 2\sqrt{b}} = \sqrt{x} + \sqrt{y}$ with $x + y = a$ , $xy = b$ .
Compare surds at the LCM of the root orders.
$a^2 - b = 1$ makes the reciprocal $a - \sqrt{b}$ .

14

Practice: 15 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.