1.5.2 Surds

What you'll learn

What surds are and why they are used to give exact answers.
How to simplify surds and perform arithmetic with them.
How to expand brackets containing surds.
How to rationalise denominators, including denominators with two terms.

Roots and exact values

A square root of a positive number is a value which, when squared, gives that number. The symbol $a\sqrt{a}$ means the positive square root of $a$ .

For example, $25=5\sqrt{25}=5$ because $5^2=25$ . Although both 5 and −5 solve $x^2=25$ , the symbol $25\sqrt{25}$ refers only to 5.

Some square roots cannot be written as exact integers or fractions. For example,

2=1.414213…\sqrt{2}=1.414213\ldots

The decimal continues without terminating or repeating.

Definition

Surd

A surd is an irrational root that is left in exact form, such as $2\sqrt{2}$ , $7\sqrt{7}$ or $53\sqrt[3]{5}$ .

At A-level, most surd questions involve square roots. Keeping a number as a surd avoids the rounding error introduced by using a decimal approximation.

Key Idea

Exact form

Unless a question asks for a decimal, leave an irrational root as a simplified surd. For example, $3\sqrt{3}$ is exact, whereas 1.732 is only an approximation.

Laws of surds

For non-negative values of $a$ and $b$ ,

ab=ab.\sqrt{a}\sqrt{b}=\sqrt{ab}.

Provided $b > 0$ ,

ab=ab.\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}.

These rules let you combine or separate roots. A particularly useful result is

aa=a.\sqrt{a}\sqrt{a}=a.

This works because multiplying a square root by itself squares it.

Example

Using the laws of surds

Simplify $6×15\sqrt{6}\times\sqrt{15}$ .

Combine the roots using $ab=ab\sqrt{a}\sqrt{b}=\sqrt{ab}$ :
$6×15=90.\sqrt{6}\times\sqrt{15}=\sqrt{90}.$
Identify a square factor of 90. Since $90=9×1090=9\times10$ ,
$90=9×10.\sqrt{90}=\sqrt{9\times10}.$
Separate the roots and evaluate the square root of 9:
$910=310.\sqrt{9}\sqrt{10}=3\sqrt{10}.$

Therefore, $6×15=310\sqrt{6}\times\sqrt{15}=3\sqrt{10}$ .

Common Mistake

Roots do not distribute over addition

In general, $a+b≠a+b\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$ . For example, $9+16=5\sqrt{9+16}=5$ , but $9+16=7\sqrt{9}+\sqrt{16}=7$ .

Simplifying surds

A surd is in simplest form when the number under the root has no square factor greater than 1.

To simplify $n\sqrt{n}$ , find the largest square number that is a factor of $n$ . Then use

ab=ab.\sqrt{ab}=\sqrt{a}\sqrt{b}.

Example

Simplifying a surd

Simplify $72\sqrt{72}$ .

Choose the largest square factor of 72. Since $72=36×272=36\times2$ ,
$72=36×2.\sqrt{72}=\sqrt{36\times2}.$
Split the root across the product:
$36×2=362.\sqrt{36\times2}=\sqrt{36}\sqrt{2}.$
Evaluate $36\sqrt{36}$ :
$72=62.\sqrt{72}=6\sqrt{2}.$

The remaining number under the root, 2, has no square factor greater than 1, so the surd is fully simplified.

Tip

Square factors

It helps to recognise the square numbers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Check the larger ones first when simplifying.

Adding and subtracting surds

Like surds have the same simplified root part. They can be collected in the same way as like algebraic terms.

For example,

35+75=105.3\sqrt{5}+7\sqrt{5}=10\sqrt{5}.

However, $2\sqrt{2}$ and $3\sqrt{3}$ are not like surds, so they cannot be combined.

Always simplify each surd before deciding whether the terms are like surds.

Example

Collecting like surds

Simplify $48+227−3\sqrt{48}+2\sqrt{27}-\sqrt{3}$ .

Simplify each root using its largest square factor:
$48=16×3=43\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}$
and
$227=29×3=63.2\sqrt{27}=2\sqrt{9\times3}=6\sqrt{3}.$
Substitute these simplified forms into the original expression:
$43+63−3.4\sqrt{3}+6\sqrt{3}-\sqrt{3}.$
Collect the coefficients of $3\sqrt{3}$ :
$(4+6−1)3=93.(4+6-1)\sqrt{3}=9\sqrt{3}.$

Common Mistake

Adding unlike surds

Do not write $2+3=5\sqrt{2}+\sqrt{3}=\sqrt{5}$ . Addition cannot be moved inside a square root, and unlike surds must remain as separate terms.

Multiplying brackets containing surds

Brackets containing surds are expanded using the same distributive rules as ordinary algebraic brackets. Multiply every term in the first bracket by every term in the second.

Remember that

(a)2=a.\left(\sqrt{a}\right)^2=a.

Example

Expanding surd brackets

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

Multiply each pair of terms:
$(3+5)(2−5)=6−35+25−(5)2.(3+\sqrt{5})(2-\sqrt{5}) =6-3\sqrt{5}+2\sqrt{5}-\left(\sqrt{5}\right)^2.$
Use $(5)2=5\left(\sqrt{5}\right)^2=5$ :
$6−35+25−5.6-3\sqrt{5}+2\sqrt{5}-5.$
Collect the rational terms and the like surds:
$1−5.1-\sqrt{5}.$

Conjugate surds

Two expressions such as $a+ba+\sqrt{b}$ and $a−ba-\sqrt{b}$ are called conjugates. They contain the same terms but have opposite signs between them.

Definition

Conjugate

The conjugate of $a+ba+\sqrt{b}$ is $a−ba-\sqrt{b}$ , and the conjugate of $a−ba-\sqrt{b}$ is $a+ba+\sqrt{b}$ .

Multiplying conjugates uses the difference of two squares:

(a+b)(a−b)=a2−(b)2=a2−b.(a+\sqrt{b})(a-\sqrt{b}) =a^2-\left(\sqrt{b}\right)^2 =a^2-b.

The surd terms cancel, leaving a rational result.

Example

Multiplying conjugates

Simplify $(4+7)(4−7)(4+\sqrt{7})(4-\sqrt{7})$ .

Recognise the brackets as conjugates, so use $x+y)(x-y)=x^2-y^2$ :
$(4+7)(4−7)=42−(7)2.(4+\sqrt{7})(4-\sqrt{7}) =4^2-\left(\sqrt{7}\right)^2.$
Evaluate the squares:
$16 - 7 = 9.$

Therefore, the product of the conjugates is 9.

Rationalising a denominator

A fraction has a rationalised denominator when its denominator contains no surds.

Definition

Rationalising the denominator

To rationalise the denominator means to rewrite a fraction in an equivalent form so that there is no irrational root in its denominator.

You multiply the numerator and denominator by a carefully chosen expression. This does not change the fraction because you are effectively multiplying it by 1.

A denominator containing one surd

If the denominator is $a\sqrt{a}$ , multiply the numerator and denominator by $a\sqrt{a}$ .

Example

Rationalising a single-surd denominator

Rationalise and simplify $53\frac{5}{\sqrt{3}}$ .

Multiply the numerator and denominator by $3\sqrt{3}$ :
$53×33=5333.\frac{5}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} =\frac{5\sqrt{3}}{\sqrt{3}\sqrt{3}}.$
Use $33=3\sqrt{3}\sqrt{3}=3$ :
$533.\frac{5\sqrt{3}}{3}.$

The denominator is now rational.

If a numerical coefficient is also present, the same method applies. For example,

325=3510.\frac{3}{2\sqrt{5}} =\frac{3\sqrt{5}}{10}.

A denominator containing two terms

If the denominator has the form $a+ba+\sqrt{b}$ or $a−ba-\sqrt{b}$ , multiply by its conjugate. The difference of two squares then removes the surd from the denominator.

Example

Rationalising a two-term denominator

Rationalise and simplify $32+5\frac{3}{2+\sqrt{5}}$ .

The conjugate of $2+52+\sqrt{5}$ is $2−52-\sqrt{5}$ . Multiply the numerator and denominator by it:
$32+5×2−52−5=3(2−5)(2+5)(2−5).\frac{3}{2+\sqrt{5}}\times\frac{2-\sqrt{5}}{2-\sqrt{5}} =\frac{3(2-\sqrt{5})}{(2+\sqrt{5})(2-\sqrt{5})}.$
Expand the numerator and use the difference of two squares in the denominator:
$6−3522−(5)2=6−354−5.\frac{6-3\sqrt{5}}{2^2-\left(\sqrt{5}\right)^2} =\frac{6-3\sqrt{5}}{4-5}.$
Simplify the denominator and then the fraction:
$6−35−1=35−6.\frac{6-3\sqrt{5}}{-1}=3\sqrt{5}-6.$

Therefore,

32+5=35−6.\frac{3}{2+\sqrt{5}}=3\sqrt{5}-6.

Common Mistake

Using the wrong conjugate

Change only the sign between the two denominator terms. The conjugate of $3−23-\sqrt{2}$ is $3+23+\sqrt{2}$ , not $−3+2-3+\sqrt{2}$ .

Tip

Check by multiplying back

After rationalising, multiply your answer by the original denominator. If you recover the original numerator, your result is consistent.

Common Mistake

A zero denominator

A denominator must never equal zero. For example, an expression with denominator $a−ba-\sqrt{b}$ is undefined if $a=ba=\sqrt{b}$ .

Equations involving surds

Surds often appear when solving quadratic equations or using exact lengths. Treat them like algebraic quantities: simplify first, collect like surds, and keep the answer exact unless a decimal is requested.

Example

Solving an equation with surds

Solve $x8+18=72x\sqrt{8}+\sqrt{18}=7\sqrt{2}$ .

Simplify each surd:
$8=22and18=32.\sqrt{8}=2\sqrt{2} \quad\text{and}\quad \sqrt{18}=3\sqrt{2}.$
Substitute and collect the terms involving $2\sqrt{2}$ :
$2x2+32=72,2x\sqrt{2}+3\sqrt{2}=7\sqrt{2},$
so
$2x2=42.2x\sqrt{2}=4\sqrt{2}.$
Divide both sides by $2\sqrt{2}$ and solve:
$\quad\Rightarrow\quad x=2.$

Exam technique

In the exam

Simplify every surd before adding, subtracting or comparing terms.
Keep answers exact unless the question explicitly requests a decimal approximation.
For a one-term surd denominator, multiply by that surd; for a two-term denominator, multiply by its conjugate.
Expand brackets carefully and use $(a)2=a\left(\sqrt{a}\right)^2=a$ whenever a root is squared.
Check that your final denominator is rational and that every remaining surd is fully simplified.

Self review

Check yourself

Can you simplify $200\sqrt{200}$ and then collect it with $323\sqrt{2}$ ?
How would you rationalise $432\frac{4}{3\sqrt{2}}$ ?
What conjugate would you use to rationalise $54−3\frac{5}{4-\sqrt{3}}$ ?

What you'll learn

What surds are and why they are used to give exact answers.
How to simplify surds and perform arithmetic with them.
How to expand brackets containing surds.
How to rationalise denominators, including denominators with two terms.

Roots and exact values

A square root of a positive number is a value which, when squared, gives that number. The symbol $a\sqrt{a}$ means the positive square root of $a$ .

For example, $25=5\sqrt{25}=5$ because $5^2=25$ . Although both 5 and −5 solve $x^2=25$ , the symbol $25\sqrt{25}$ refers only to 5.

Some square roots cannot be written as exact integers or fractions. For example,

2=1.414213…\sqrt{2}=1.414213\ldots

The decimal continues without terminating or repeating.

Definition

Surd

A surd is an irrational root that is left in exact form, such as $2\sqrt{2}$ , $7\sqrt{7}$ or $53\sqrt[3]{5}$ .

At A-level, most surd questions involve square roots. Keeping a number as a surd avoids the rounding error introduced by using a decimal approximation.

Key Idea

Exact form

Unless a question asks for a decimal, leave an irrational root as a simplified surd. For example, $3\sqrt{3}$ is exact, whereas 1.732 is only an approximation.

Laws of surds

For non-negative values of $a$ and $b$ ,

ab=ab.\sqrt{a}\sqrt{b}=\sqrt{ab}.

Provided $b > 0$ ,

ab=ab.\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}.

These rules let you combine or separate roots. A particularly useful result is

aa=a.\sqrt{a}\sqrt{a}=a.

This works because multiplying a square root by itself squares it.

Example

Using the laws of surds

Simplify $6×15\sqrt{6}\times\sqrt{15}$ .

Combine the roots using $ab=ab\sqrt{a}\sqrt{b}=\sqrt{ab}$ :
$6×15=90.\sqrt{6}\times\sqrt{15}=\sqrt{90}.$
Identify a square factor of 90. Since $90=9×1090=9\times10$ ,
$90=9×10.\sqrt{90}=\sqrt{9\times10}.$
Separate the roots and evaluate the square root of 9:
$910=310.\sqrt{9}\sqrt{10}=3\sqrt{10}.$

Therefore, $6×15=310\sqrt{6}\times\sqrt{15}=3\sqrt{10}$ .

Common Mistake

Roots do not distribute over addition

In general, $a+b≠a+b\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$ . For example, $9+16=5\sqrt{9+16}=5$ , but $9+16=7\sqrt{9}+\sqrt{16}=7$ .

Simplifying surds

A surd is in simplest form when the number under the root has no square factor greater than 1.

To simplify $n\sqrt{n}$ , find the largest square number that is a factor of $n$ . Then use

ab=ab.\sqrt{ab}=\sqrt{a}\sqrt{b}.

Example

Simplifying a surd

Simplify $72\sqrt{72}$ .

Choose the largest square factor of 72. Since $72=36×272=36\times2$ ,
$72=36×2.\sqrt{72}=\sqrt{36\times2}.$
Split the root across the product:
$36×2=362.\sqrt{36\times2}=\sqrt{36}\sqrt{2}.$
Evaluate $36\sqrt{36}$ :
$72=62.\sqrt{72}=6\sqrt{2}.$

The remaining number under the root, 2, has no square factor greater than 1, so the surd is fully simplified.

Tip

Square factors

It helps to recognise the square numbers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Check the larger ones first when simplifying.

Adding and subtracting surds

Like surds have the same simplified root part. They can be collected in the same way as like algebraic terms.

For example,

35+75=105.3\sqrt{5}+7\sqrt{5}=10\sqrt{5}.

However, $2\sqrt{2}$ and $3\sqrt{3}$ are not like surds, so they cannot be combined.

Always simplify each surd before deciding whether the terms are like surds.

Example

Collecting like surds

Simplify $48+227−3\sqrt{48}+2\sqrt{27}-\sqrt{3}$ .

Simplify each root using its largest square factor:
$48=16×3=43\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}$
and
$227=29×3=63.2\sqrt{27}=2\sqrt{9\times3}=6\sqrt{3}.$
Substitute these simplified forms into the original expression:
$43+63−3.4\sqrt{3}+6\sqrt{3}-\sqrt{3}.$
Collect the coefficients of $3\sqrt{3}$ :
$(4+6−1)3=93.(4+6-1)\sqrt{3}=9\sqrt{3}.$

Common Mistake

Adding unlike surds

Do not write $2+3=5\sqrt{2}+\sqrt{3}=\sqrt{5}$ . Addition cannot be moved inside a square root, and unlike surds must remain as separate terms.

Multiplying brackets containing surds

Brackets containing surds are expanded using the same distributive rules as ordinary algebraic brackets. Multiply every term in the first bracket by every term in the second.

Remember that

(a)2=a.\left(\sqrt{a}\right)^2=a.

Example

Expanding surd brackets

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

Multiply each pair of terms:
$(3+5)(2−5)=6−35+25−(5)2.(3+\sqrt{5})(2-\sqrt{5}) =6-3\sqrt{5}+2\sqrt{5}-\left(\sqrt{5}\right)^2.$
Use $(5)2=5\left(\sqrt{5}\right)^2=5$ :
$6−35+25−5.6-3\sqrt{5}+2\sqrt{5}-5.$
Collect the rational terms and the like surds:
$1−5.1-\sqrt{5}.$

Conjugate surds

Two expressions such as $a+ba+\sqrt{b}$ and $a−ba-\sqrt{b}$ are called conjugates. They contain the same terms but have opposite signs between them.

Definition

Conjugate

The conjugate of $a+ba+\sqrt{b}$ is $a−ba-\sqrt{b}$ , and the conjugate of $a−ba-\sqrt{b}$ is $a+ba+\sqrt{b}$ .

Multiplying conjugates uses the difference of two squares:

(a+b)(a−b)=a2−(b)2=a2−b.(a+\sqrt{b})(a-\sqrt{b}) =a^2-\left(\sqrt{b}\right)^2 =a^2-b.

The surd terms cancel, leaving a rational result.

Example

Multiplying conjugates

Simplify $(4+7)(4−7)(4+\sqrt{7})(4-\sqrt{7})$ .

Recognise the brackets as conjugates, so use $x+y)(x-y)=x^2-y^2$ :
$(4+7)(4−7)=42−(7)2.(4+\sqrt{7})(4-\sqrt{7}) =4^2-\left(\sqrt{7}\right)^2.$
Evaluate the squares:
$16 - 7 = 9.$

Therefore, the product of the conjugates is 9.

Rationalising a denominator

A fraction has a rationalised denominator when its denominator contains no surds.

Definition

Rationalising the denominator

To rationalise the denominator means to rewrite a fraction in an equivalent form so that there is no irrational root in its denominator.

You multiply the numerator and denominator by a carefully chosen expression. This does not change the fraction because you are effectively multiplying it by 1.

A denominator containing one surd

If the denominator is $a\sqrt{a}$ , multiply the numerator and denominator by $a\sqrt{a}$ .

Example

Rationalising a single-surd denominator

Rationalise and simplify $53\frac{5}{\sqrt{3}}$ .

Multiply the numerator and denominator by $3\sqrt{3}$ :
$53×33=5333.\frac{5}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} =\frac{5\sqrt{3}}{\sqrt{3}\sqrt{3}}.$
Use $33=3\sqrt{3}\sqrt{3}=3$ :
$533.\frac{5\sqrt{3}}{3}.$

The denominator is now rational.

If a numerical coefficient is also present, the same method applies. For example,

325=3510.\frac{3}{2\sqrt{5}} =\frac{3\sqrt{5}}{10}.

A denominator containing two terms

If the denominator has the form $a+ba+\sqrt{b}$ or $a−ba-\sqrt{b}$ , multiply by its conjugate. The difference of two squares then removes the surd from the denominator.

Example

Rationalising a two-term denominator

Rationalise and simplify $32+5\frac{3}{2+\sqrt{5}}$ .

The conjugate of $2+52+\sqrt{5}$ is $2−52-\sqrt{5}$ . Multiply the numerator and denominator by it:
$32+5×2−52−5=3(2−5)(2+5)(2−5).\frac{3}{2+\sqrt{5}}\times\frac{2-\sqrt{5}}{2-\sqrt{5}} =\frac{3(2-\sqrt{5})}{(2+\sqrt{5})(2-\sqrt{5})}.$
Expand the numerator and use the difference of two squares in the denominator:
$6−3522−(5)2=6−354−5.\frac{6-3\sqrt{5}}{2^2-\left(\sqrt{5}\right)^2} =\frac{6-3\sqrt{5}}{4-5}.$
Simplify the denominator and then the fraction:
$6−35−1=35−6.\frac{6-3\sqrt{5}}{-1}=3\sqrt{5}-6.$

Therefore,

32+5=35−6.\frac{3}{2+\sqrt{5}}=3\sqrt{5}-6.

Common Mistake

Using the wrong conjugate

Change only the sign between the two denominator terms. The conjugate of $3−23-\sqrt{2}$ is $3+23+\sqrt{2}$ , not $−3+2-3+\sqrt{2}$ .

Tip

Check by multiplying back

After rationalising, multiply your answer by the original denominator. If you recover the original numerator, your result is consistent.

Common Mistake

A zero denominator

A denominator must never equal zero. For example, an expression with denominator $a−ba-\sqrt{b}$ is undefined if $a=ba=\sqrt{b}$ .

Equations involving surds

Example

Solving an equation with surds

Solve $x8+18=72x\sqrt{8}+\sqrt{18}=7\sqrt{2}$ .

Simplify each surd:
$8=22and18=32.\sqrt{8}=2\sqrt{2} \quad\text{and}\quad \sqrt{18}=3\sqrt{2}.$
Substitute and collect the terms involving $2\sqrt{2}$ :
$2x2+32=72,2x\sqrt{2}+3\sqrt{2}=7\sqrt{2},$
so
$2x2=42.2x\sqrt{2}=4\sqrt{2}.$
Divide both sides by $2\sqrt{2}$ and solve:
$\quad\Rightarrow\quad x=2.$

Exam technique

In the exam

Simplify every surd before adding, subtracting or comparing terms.
Keep answers exact unless the question explicitly requests a decimal approximation.
For a one-term surd denominator, multiply by that surd; for a two-term denominator, multiply by its conjugate.
Expand brackets carefully and use $(a)2=a\left(\sqrt{a}\right)^2=a$ whenever a root is squared.
Check that your final denominator is rational and that every remaining surd is fully simplified.

Self review

Check yourself

Can you simplify $200\sqrt{200}$ and then collect it with $323\sqrt{2}$ ?
How would you rationalise $432\frac{4}{3\sqrt{2}}$ ?
What conjugate would you use to rationalise $54−3\frac{5}{4-\sqrt{3}}$ ?

What you'll learn

Roots and exact values

Laws of surds

Simplifying surds

Adding and subtracting surds

Multiplying brackets containing surds

Conjugate surds

Rationalising a denominator

A denominator containing one surd

A denominator containing two terms

Equations involving surds

Review 1.5.2 Surds by teaching Genie

1.5.2 Surds

What you'll learn

Roots and exact values

Laws of surds

Simplifying surds

Adding and subtracting surds

Multiplying brackets containing surds

Conjugate surds

Rationalising a denominator

A denominator containing one surd

A denominator containing two terms

Equations involving surds

Review 1.5.2 Surds by teaching Genie

1.5.2 Surds

What you'll learn

Roots and exact values

Laws of surds

Simplifying surds

Adding and subtracting surds

Multiplying brackets containing surds

Conjugate surds

Rationalising a denominator

A denominator containing one surd

A denominator containing two terms

Equations involving surds

Review 1.5.2 Surds by teaching Genie

1.5.2 Surds Revision Guide

1.5.2 Surds

What you'll learn

Roots and exact values

Laws of surds

Simplifying surds

Adding and subtracting surds

Multiplying brackets containing surds

Conjugate surds

Rationalising a denominator

A denominator containing one surd

A denominator containing two terms

Equations involving surds

Review 1.5.2 Surds by teaching Genie

1.5.2 Surds Revision Guide