Surds: rationalisation and square roots of surds — notes, question types & practice · Algebra

01

Overview

To clear a root from a denominator, multiply top and bottom by the conjugate, the same terms with the middle sign flipped:

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

So $\dfrac{1}{\sqrt{7}-\sqrt{6}} = \sqrt{7}+\sqrt{6}$ , because the denominator product is $7 - 6 = 1$ . With a coefficient, divide once: $\dfrac{4}{\sqrt{5}-1} = \dfrac{4(\sqrt{5}+1)}{4} = \sqrt{5}+1$ , and $\dfrac{3}{\sqrt{5}+\sqrt{2}} = \sqrt{5}-\sqrt{2}$ .

02

Denesting a nested root

Try $\sqrt{a + b\sqrt{c}} = \sqrt{p} + \sqrt{q}$ . Squaring gives the two conditions: $p + q = a$ and $2\sqrt{pq} = b\sqrt{c}$ . Examples: $\sqrt{15 + 4\sqrt{14}} = 2\sqrt{2} + \sqrt{7}$ , since $8 + 7 = 15$ and $2\sqrt{56} = 4\sqrt{14}$ ; and $\sqrt{11 + 6\sqrt{2}} = 3 + \sqrt{2}$ , since $9 + 2 = 11$ and $2\sqrt{9} = 6$ .

Rule: Force the inner coefficient to the shape $2\sqrt{\cdot}$ first: $6\sqrt{2}$ is $2\sqrt{18}$ , so the product you need is 18, not 2.

03

The product-one conjugate pair

When $x = p + \sqrt{q}$ and $p^2 - q = 1$ , the reciprocal is the conjugate: $\dfrac{1}{x} = p - \sqrt{q}$ . Then the reciprocal ladder runs free. With $x = 2+\sqrt{3}$ : $x + \dfrac{1}{x} = 4$ , the square rung gives $14$ , and the cube rung gives $52$ . With $x = 5+2\sqrt{6}$ : $25 - 24 = 1$ , so $x + \dfrac{1}{x} = 10$ and the square rung gives $98$ .

Watch: Check $p^2 - q$ before claiming the reciprocal. If it is not 1 or a perfect square, do the full rationalisation instead.

04

Telescoping chains

Each term $\dfrac{1}{\sqrt{n+1}+\sqrt{n}}$ equals $\sqrt{n+1}-\sqrt{n}$ , so a chain cancels in the middle:

\dfrac{1}{\sqrt{5}+\sqrt{4}} + \dfrac{1}{\sqrt{6}+\sqrt{5}} + \dfrac{1}{\sqrt{7}+\sqrt{6}} = \sqrt{7} - \sqrt{4} = \sqrt{7}-2

Write the answer as last root minus first root and skip the middle terms entirely. A chain running from $\sqrt{3}$ to $\sqrt{10}$ collapses to $\sqrt{10}-\sqrt{3}$ the same way, however many terms it lists.

05

Comparing surds

Raise every surd to the LCM of the root orders and compare plain integers. For $\sqrt{2}$ , $\sqrt[3]{3}$ , $\sqrt[4]{5}$ : orders 2, 3, 4 give LCM 12, and twelfth powers are $2^6 = 64$ , $3^4 = 81$ , $5^3 = 125$ . The last is largest, so $\sqrt[4]{5}$ wins. For differences of roots, use $\sqrt{a}-\sqrt{b} = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}$ : with equal numerators, the pair with larger roots gives the smaller difference. That is why $\sqrt{5}-\sqrt{4}$ beats $\sqrt{7}-\sqrt{6}$ .

Tip: Keep three decimal values ready: $\sqrt{2} \approx 1.414$ , $\sqrt{3} \approx 1.732$ , $\sqrt{5} \approx 2.236$ . They settle close comparisons in seconds.

06

Question types you will see

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

Type 1common2 practice Q

Denesting a nested surd

How to spot it:

A square root wrapped around a number plus or minus a root, to be written without nesting.

\sqrt{a+2\sqrt{b}} = \sqrt{m}+\sqrt{n},\ m+n=a,\ mn=b

Method

Rewrite the inner term in the $2\sqrt{\cdot}$ shape.
Find two numbers with sum $a$ and product $b$ .
Answer is $\sqrt{m}+\sqrt{n}$ , or the difference for the minus form.
Re-square mentally to confirm.

Why it works:

Squaring $\sqrt{m}+\sqrt{n}$ returns exactly $a + 2\sqrt{b}$ .

Try this

Simplify $\sqrt{16 - 6\sqrt{7}}$ .

Show solution

$6\sqrt{7} = 2\sqrt{63}$ , so sum $16$ , product $63$ .
The pair is $9$ and $7$ .
$\sqrt{16-6\sqrt{7}} = \sqrt{9}-\sqrt{7} = 3-\sqrt{7}$ .

Answer

3 - sqrt(7)

Type 2very common3 practice Q

Rationalise the denominator

How to spot it:

A fraction with a root or a binomial surd in the denominator.

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

Method

Multiply top and bottom by the conjugate.
The denominator becomes $a - b$ .
Cancel any common factor before expanding.

Why it works:

The conjugate product removes every root from the denominator.

Try this

Find the simplified value of $\dfrac{6}{\sqrt{7}-1}$ .

Show solution

Multiply by $\dfrac{\sqrt{7}+1}{\sqrt{7}+1}$ .
Denominator: $7 - 1 = 6$ .
$\dfrac{6(\sqrt{7}+1)}{6} = \sqrt{7}+1$ .

Answer

sqrt(7) + 1

Type 3very common4 practice Q

Conjugate pair with product one

How to spot it:

$x = p + \sqrt{q}$ is given, and $x + \frac{1}{x}$ or a higher rung is asked.

p^2 - q = 1 \Rightarrow \dfrac{1}{x} = p - \sqrt{q},\ x + \dfrac{1}{x} = 2p

Method

Test $p^2 - q$ ; confirm it equals 1.
Write the reciprocal as the conjugate.
Add to get $x + \frac{1}{x} = 2p$ .
Climb the reciprocal ladder if a higher rung is asked.

Why it works:

A product-one pair means the conjugate is exactly the reciprocal.

Try this

If $x = 5 + 2\sqrt{6}$ , find $x + \dfrac{1}{x}$ .

Show solution

$p^2 - q = 25 - 24 = 1$ .
$\dfrac{1}{x} = 5 - 2\sqrt{6}$ .
$x + \dfrac{1}{x} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$ .

Answer

10

Type 4common2 practice Q

Comparing surds of different orders

How to spot it:

A list of unlike roots asks which is greatest or smallest.

\text{raise all to the LCM of the root orders}

Method

Take the LCM of all root orders.
Raise each surd to that power; roots vanish.
Compare the integers.

Why it works:

A common power turns every surd into a whole number.

Try this

Which is the greatest: $\sqrt{2}$ , $\sqrt[3]{3}$ or $\sqrt[4]{5}$ ?

Show solution

Orders 2, 3, 4 give LCM 12.
Twelfth powers: $2^6 = 64$ , $3^4 = 81$ , $5^3 = 125$ .
$125$ is largest, so $\sqrt[4]{5}$ is the greatest.

Answer

fourth root of 5

Type 5common

Ranking fractions with root-sum denominators

How to spot it:

Options like $\dfrac{1}{\sqrt{n}+\sqrt{n+1}}$ ask which fraction is largest or smallest.

\dfrac{1}{\sqrt{a}+\sqrt{b}} = \dfrac{\sqrt{a}-\sqrt{b}}{a-b}

Method

Rationalise every option with its conjugate.
Each difference of radicands is $1$ here, so each value is a plain root difference.
With equal numerators, the smallest pair of roots gives the largest value.
Rank and pick.

Why it works:

After rationalisation all options share numerator $1$ , so only the size of the root pair decides.

Try this

Which is the largest: $\dfrac{1}{\sqrt{3}+\sqrt{2}}$ , $\dfrac{1}{\sqrt{5}+\sqrt{4}}$ or $\dfrac{1}{\sqrt{7}+\sqrt{6}}$ ?

Show solution

Rationalising gives $\sqrt{3}-\sqrt{2}$ , $\sqrt{5}-2$ and $\sqrt{7}-\sqrt{6}$ .
All equal $\dfrac{1}{\text{sum of the roots}}$ , so the smallest roots win.
$\sqrt{3}+\sqrt{2}$ is the smallest sum, so $\sqrt{3}-\sqrt{2}$ is the largest value.

Answer

1/(sqrt(3)+sqrt(2))

07

Formula sheet

Rationalisation

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

Conjugate product

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

Root of a surd, plus

\sqrt{a+2\sqrt{b}} = \sqrt{m}+\sqrt{n},\ m+n = a,\ mn = b

Root of a surd, minus

\sqrt{a-2\sqrt{b}} = \sqrt{m}-\sqrt{n}\ \ (m > n)

Product-one pair

x = p+\sqrt{q},\ p^2-q = 1 \Rightarrow \tfrac1x = p-\sqrt{q},\ x+\tfrac1x = 2p

Telescoping sum

\sum \frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{\text{last}} - \sqrt{\text{first}}

Difference of roots

\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}

08

Shortcuts that save time

⚡ Denest by sum and product

Rewrite the inner coefficient as 2 root something, then find two numbers with the given sum and product.

Example

Simplify $\sqrt{17 - 4\sqrt{13}}$ .

Show solution

$4\sqrt{13} = 2\sqrt{52}$ , so sum $17$ , product $52$ .
The pair is $13$ and $4$ .
$\sqrt{17-4\sqrt{13}} = \sqrt{13} - 2$ .

Answer

sqrt(13) - 2

⚡ Spot the product-one conjugate

For x = p + sqrt(q), test p squared minus q. If it is 1, the reciprocal is the conjugate and the ladder runs.

Example

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

Show solution

$p^2 - q = 5 - 4 = 1$ , so $\dfrac{1}{x} = \sqrt{5} - 2$ .
$x - \dfrac{1}{x} = 4$ .
$x^2+\dfrac{1}{x^2} = 4^2 + 2 = 18$ .

Answer

18

⚡ Telescoping rationalisation

Each fraction over consecutive roots becomes a difference; middle terms cancel, leaving last root minus first.

Example

Find $\dfrac{1}{\sqrt{4}+\sqrt{5}} + \dfrac{1}{\sqrt{5}+\sqrt{6}} + \dfrac{1}{\sqrt{6}+\sqrt{7}}$ .

Show solution

Each term is $\sqrt{n+1} - \sqrt{n}$ .
Middle roots cancel pairwise.
Answer $= \sqrt{7} - \sqrt{4} = \sqrt{7} - 2$ .

Answer

sqrt(7) - 2

09

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Writing $\sqrt{a} + \sqrt{b} = \sqrt{a+b}$ .

Roots do not split over sums; only products split under a root.

Mistake 02

Denesting without making the inner coefficient 2.

Rewrite $b\sqrt{c}$ as $2\sqrt{b^2c/4}$ first, so the product is read correctly.

Mistake 03

Putting the smaller root first in $\sqrt{a - 2\sqrt{b}}$ .

The minus form is larger root minus smaller root, a positive value.

Mistake 04

Claiming the reciprocal is the conjugate without checking $p^2 - q$ .

Test $p^2 - q$ ; only 1 or a perfect square gives the shortcut.

Mistake 05

Comparing surds of different orders by their radicands.

Raise all to the LCM of the root orders, then compare integers.

10

Quick revision

Read this the night before the exam.

Conjugate rationalisation divides by $a - b$ .
$\sqrt{a+2\sqrt{b}} = \sqrt{m}+\sqrt{n}$ with $m+n = a$ , $mn = b$ .
$p^2 - q = 1$ makes the reciprocal $p - \sqrt{q}$ .
Chains of $\dfrac{1}{\sqrt{n}+\sqrt{n+1}}$ collapse to last minus first.
Compare at the LCM of root orders.
$\sqrt{2} \approx 1.414$ , $\sqrt{3} \approx 1.732$ , $\sqrt{5} \approx 2.236$ .

11

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.