Surds [H]

This lesson covers simplifying surds, adding and subtracting surds, multiplying surds, expanding brackets involving surds, and rationalising the denominator. Surds is a Higher-tier only topic in the Edexcel GCSE Mathematics (1MA1) specification and frequently appears on Papers 1, 2 and 3.

Key Vocabulary

Term	Meaning
Surd	An irrational root that cannot be simplified to a whole number, e.g. $\sqrt{2}$ , $\sqrt{5}$
Rational number	A number that can be written as a fraction p/q where p, q are integers and $q \neq 0$
Irrational number	A number that cannot be written as a fraction — its decimal goes on forever without repeating
Rationalise the denominator	Rewrite a fraction so that the denominator contains no surds
Conjugate	The expression formed by changing the sign between two terms, e.g. the conjugate of $(3 + \sqrt{2})$ is $(3 - \sqrt{2})$

What Is a Surd?

A surd is a root that cannot be simplified to a rational number.

$\sqrt{4} = 2$ — this is NOT a surd (it simplifies to a whole number)
$\sqrt{5} = 2.2360679$ ... — this IS a surd (it cannot be simplified)

Key Fact: $\sqrt{n}$ is a surd if n is not a perfect square.

Simplifying Surds

To simplify a surd, find the largest square number that is a factor of the number under the root.

Rule: $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

Worked Example 1

Simplify $\sqrt{50}$

Find the largest square factor of 50: $50 = 25 \times 2$
$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$

Answer: $5\sqrt{2}$

Worked Example 2

Simplify $\sqrt{72}$

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

Answer: $6\sqrt{2}$

Worked Example 3

Simplify $\sqrt{200}$

$200 = 100 \times 2$
$\sqrt{200} = \sqrt{100} \times \sqrt{2} = 10\sqrt{2}$

Answer: $10\sqrt{2}$

Edexcel Exam Tip: Always use the LARGEST square factor to simplify in one step. Using smaller factors (e.g. $\sqrt{72} = \sqrt{4} \times \sqrt{18} = 2\sqrt{18})$ means you'll need to simplify again. Examiners want the fully simplified form.

Adding and Subtracting Surds

You can only add or subtract surds with the same number under the root (like collecting "like terms").

Worked Example 4

Simplify $3\sqrt{5} + 7\sqrt{5}$

$3\sqrt{5} + 7\sqrt{5} = 10\sqrt{5} ($ just like 3x + 7x = 10x)

Worked Example 5

Simplify $5\sqrt{3} - 2\sqrt{3} + 4\sqrt{3}$

$= (5 - 2 + 4)\sqrt{3} = 7\sqrt{3}$

Worked Example 6

Simplify $\sqrt{12} + \sqrt{27}$

First, simplify each surd:

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

Now add: $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$

Answer: $5\sqrt{3}$

Multiplying Surds

Rule: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

Worked Example 7

Simplify $\sqrt{3} \times \sqrt{7}$

$\sqrt{3} \times \sqrt{7} = \sqrt{21}$

Worked Example 8

Simplify $2\sqrt{5} \times 3\sqrt{2}$

$= (2 \times 3) \times (\sqrt{5} \times \sqrt{2}) = 6\sqrt{10}$

Worked Example 9

Simplify $(\sqrt{6})^{2}$

$(\sqrt{6})^{2} = \sqrt{6} \times \sqrt{6} = \sqrt{36} = 6$

Key Rule: $(\sqrt{a})^{2} = a$ . This is used extensively in rationalising and expanding.

Expanding Brackets with Surds

Use the same techniques as algebraic expansion.

Worked Example 10

Expand and simplify $(2 + \sqrt{3})(4 - \sqrt{3})$

Use FOIL (or grid method):

$= 2 \times 4 + 2 \times (-\sqrt{3}) + \sqrt{3} \times 4 + \sqrt{3} \times (-\sqrt{3})$

$= 8 - 2\sqrt{3} + 4\sqrt{3} - 3$

$= 5 + 2\sqrt{3}$

Answer: $5 + 2\sqrt{3}$

Worked Example 11

Expand and simplify $(3 + \sqrt{5})^{2}$

$(3 + \sqrt{5})^{2} = (3 + \sqrt{5})(3 + \sqrt{5})$

$= 9 + 3\sqrt{5} + 3\sqrt{5} + 5$

$= 14 + 6\sqrt{5}$

Answer: $14 + 6\sqrt{5}$

Worked Example 12

Expand $(\sqrt{7} + 2)(\sqrt{7} - 2)$

This is the difference of two squares: $(a + b)(a - b) = a^{2} - b^{2}$

$= (\sqrt{7})^{2} - 2^{2} = 7 - 4 = 3$

Answer: 3

Important: When you multiply conjugate pairs, the surds cancel out, leaving a rational number. This is the basis for rationalising the denominator.

Rationalising the Denominator

"Rationalising the denominator" means rewriting a fraction so the denominator is a rational number (no surds).

Case 1: Denominator is a simple $surd (\sqrt{a})$

Multiply top and bottom by $\sqrt{a}$ .

Worked Example 13

Rationalise $5/\sqrt{3}$

$= (5 \times \sqrt{3}) / (\sqrt{3} \times \sqrt{3}) = 5\sqrt{3} / 3$

Answer: $5\sqrt{3}/3$

Worked Example 14

Rationalise $8/(3\sqrt{2})$

$= (8 \times \sqrt{2}) / (3\sqrt{2} \times \sqrt{2}) = 8\sqrt{2} / 6 = 4\sqrt{2}/3$

Answer: $4\sqrt{2}/3$

Case 2: Denominator is of the form $(a \pm \sqrt{b})$

Multiply top and bottom by the conjugate (change the sign).

Worked Example 15

Rationalise $6/(3 + \sqrt{2})$

Multiply by $(3 - \sqrt{2})/(3 - \sqrt{2})$ :

Numerator: $6(3 - \sqrt{2}) = 18 - 6\sqrt{2}$

Denominator: $(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$

Answer: $(18 - 6\sqrt{2})/7$

Worked Example 16

Rationalise $1/(\sqrt{5} - \sqrt{3})$

Multiply by $(\sqrt{5} + \sqrt{3})/(\sqrt{5} + \sqrt{3})$ :

Numerator: $1 \times (\sqrt{5} + \sqrt{3}) = \sqrt{5} + \sqrt{3}$

Denominator: $(\sqrt{5})^{2} - (\sqrt{3})^{2} = 5 - 3 = 2$

Answer: $(\sqrt{5} + \sqrt{3})/2$

Common Mistakes and Misconceptions

Mistake	Correction
$\sqrt{9} + \sqrt{16} = \sqrt{25} = 5$	$\sqrt{9} + \sqrt{16} = 3 + 4 = 7 ($ you cannot add numbers under roots)
$\sqrt{a + b} = \sqrt{a} + \sqrt{b}$	WRONG — you can only split roots over multiplication, not addition
$\sqrt{50} = \sqrt{5} \times \sqrt{10}$ is "simplified"	Must use largest square factor: $\sqrt{50} = 5\sqrt{2}$
Rationalising by multiplying denominator only	You must multiply BOTH numerator AND denominator
$(\sqrt{3})^{2} = \sqrt{9}$	$(\sqrt{3})^{2} = 3 ($ not $\sqrt{9})$ . This is correct either way since $\sqrt{9} = 3$ , but think of it as $(\sqrt{3})^{2} = 3$

Practice Problems

Simplify $\sqrt{98}$ .
Simplify $\sqrt{45} + \sqrt{80}$ .
Simplify $2\sqrt{3} \times 4\sqrt{3}$ .
Expand and simplify $(5 - \sqrt{2})(3 + \sqrt{2})$ .
Expand and simplify $(1 + \sqrt{6})^{2}$ .
Rationalise $10/\sqrt{5}$ .
Rationalise $4/(\sqrt{7} - \sqrt{3})$ .
Show that $(2 + \sqrt{3})(2 - \sqrt{3})$ is rational.

Answers

$\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$
$\sqrt{45} = 3\sqrt{5}$ , $\sqrt{80} = 4\sqrt{5}$ . Sum $= 7\sqrt{5}$
$2 \times 4 \times \sqrt{3} \times \sqrt{3} = 8 \times 3 = 24$
$15 + 5\sqrt{2} - 3\sqrt{2} - 2 = 13 + 2\sqrt{2}$
$1 + 2\sqrt{6} + 6 = 7 + 2\sqrt{6}$
$(10\sqrt{5})/(\sqrt{5} \times \sqrt{5}) = 10\sqrt{5}/5 = 2\sqrt{5}$
Multiply by $(\sqrt{7} + \sqrt{3})/(\sqrt{7} + \sqrt{3})$ : $4(\sqrt{7} + \sqrt{3})/(7 - 3) = 4(\sqrt{7} + \sqrt{3})/4 = \sqrt{7} + \sqrt{3}$
$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$ , which is rational ✓

Surds [H]

Surds [H]

Key Vocabulary

What Is a Surd?

Simplifying Surds

Worked Example 1

Worked Example 2

Worked Example 3

Adding and Subtracting Surds

Worked Example 4

Worked Example 5

Worked Example 6

Multiplying Surds

Worked Example 7

Worked Example 8

Worked Example 9

Expanding Brackets with Surds

Worked Example 10

Worked Example 11

Worked Example 12

Rationalising the Denominator

Case 1: Denominator is a simple $surd (\sqrt{a})$

Worked Example 13

Worked Example 14

Case 2: Denominator is of the form $(a \pm \sqrt{b})$

Worked Example 15

Worked Example 16

Common Mistakes and Misconceptions

Practice Problems

Answers

Summary

More in Mathematics

Surds [H]

Surds [H]

Key Vocabulary

What Is a Surd?

Simplifying Surds

Worked Example 1

Worked Example 2

Worked Example 3

Adding and Subtracting Surds

Worked Example 4

Worked Example 5

Worked Example 6

Multiplying Surds

Worked Example 7

Worked Example 8

Worked Example 9

Expanding Brackets with Surds

Worked Example 10

Worked Example 11

Worked Example 12

Rationalising the Denominator

Case 1: Denominator is a simple surd(a)surd (\sqrt{a})surd(a​)

Worked Example 13

Worked Example 14

Case 2: Denominator is of the form (a±b)(a \pm \sqrt{b})(a±b​)

Worked Example 15

Worked Example 16

Common Mistakes and Misconceptions

Practice Problems

Answers

Summary

More in Mathematics

Case 1: Denominator is a simple $surd (\sqrt{a})$

Case 2: Denominator is of the form $(a \pm \sqrt{b})$