Surds & irrational arithmetic - Year selective Mathematics

What you will learn

recognise when a square root is already in simplest surd form,
simplify $\sqrt{n}$ by pulling out perfect-square factors,
add and subtract like surds (same radical),
multiply surds and expand brackets containing them,
rationalise a denominator of the form $\dfrac{a}{\sqrt{b}}$ .

Where surds come from — geometry:

A square with side 1 has a diagonal of exactly $\sqrt{2}$ — by Pythagoras, $\sqrt{1^2 + 1^2} = \sqrt{2}$ . This is a real length that exists in the physical world. Surds aren’t abstract — they are the exact way to name lengths that decimals can only approximate.

1. What is a surd?

A surd is a root that cannot be simplified to a rational number - $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}$ are surds; $\sqrt{4} = 2$ is not.

Core surd rule

\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}, \qquad \sqrt{\tfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}.

2. Simplifying a surd

Strategy: look for the largest perfect-square factor of the number under the root and pull it out.

Worked example 1 Simplify a surd

Simplify $\;\sqrt{72}$ .

$72 = 36 \times 2$ and $36$ is a perfect square.

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

Worked example 2 Pulling out coefficients

Simplify $\;3\sqrt{50}$ .

$50 = 25 \times 2$ , so $\sqrt{50} = 5\sqrt{2}$ , and

3\sqrt{50} = 3 \times 5\sqrt{2} = 15\sqrt{2}.

3. Adding and subtracting surds

Only like surds (same radical) combine - just like only like algebraic terms combine.

Worked example 3 Like surds

Simplify $\;5\sqrt{3} + 2\sqrt{3} - \sqrt{3}$ .

(5 + 2 - 1)\sqrt{3} = 6\sqrt{3}.

Worked example 4 Simplify first, then collect

Simplify $\;\sqrt{50} + \sqrt{18}$ .

Each surd simplifies first: $\sqrt{50} = 5\sqrt{2}$ , $\sqrt{18} = 3\sqrt{2}$ . Now they are like surds:

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

4. Multiplying surds

Multiplying surds

Roots multiply

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

Coefficients multiply separately

p\sqrt{a} \times q\sqrt{b} = pq\sqrt{ab}.

A surd squared

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

Worked example 5 Product of surds

Simplify $\;2\sqrt{3} \times 4\sqrt{6}$ .

(2 \times 4)\sqrt{3 \times 6} = 8\sqrt{18} = 8 \times 3\sqrt{2} = 24\sqrt{2}.

Worked example 6 Expand brackets

Expand $\;\sqrt{2}(3 + \sqrt{8})$ .

3\sqrt{2} + \sqrt{2}\sqrt{8} = 3\sqrt{2} + \sqrt{16} = 3\sqrt{2} + 4.

5. Rationalising a denominator

A denominator containing a surd is usually rewritten so the denominator becomes rational. Multiply top and bottom by the surd.

Single-surd rationalising

\dfrac{a}{\sqrt{b}} = \dfrac{a}{\sqrt{b}} \times \dfrac{\sqrt{b}}{\sqrt{b}} = \dfrac{a\sqrt{b}}{b}.

Worked example 7 Rationalise

Rewrite $\;\dfrac{6}{\sqrt{2}}$ with a rational denominator.

\dfrac{6}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2}.

Practice

Fluency

Simplify

Simplify $\sqrt{12}$ .
Simplify $\sqrt{27}$ .
Simplify $\sqrt{45}$ .
Simplify $\sqrt{98}$ .
Simplify $2\sqrt{32}$ .
Simplify $\sqrt{200}$ .

Fluency

Add & subtract

$\sqrt{5} + 3\sqrt{5}$ .
$7\sqrt{2} - 4\sqrt{2}$ .
$\sqrt{8} + \sqrt{18}$ .
$\sqrt{50} - \sqrt{32}$ .
$\sqrt{75} + \sqrt{27}$ .

Fluency

Multiply

$\sqrt{3} \times \sqrt{7}$ .
$2\sqrt{5} \times 3\sqrt{2}$ .
$(\sqrt{6})^2$ .
$\sqrt{8} \times \sqrt{2}$ .
$\sqrt{3}(2 + \sqrt{12})$ .

Fluency

Rationalise

Rationalise $\dfrac{1}{\sqrt{3}}$ .
Rationalise $\dfrac{10}{\sqrt{5}}$ .
Rationalise $\dfrac{\sqrt{2}}{\sqrt{8}}$ .

Reasoning

Mixed & apply

A square has area $48\text{ cm}^2$ . Find its side length in simplest surd form.
Find the exact length of the diagonal of a $3 \times 5$ rectangle in simplest surd form.
Simplify $\sqrt{18} + \sqrt{8} - \sqrt{50}$ .
Show that $\sqrt{2} \times \sqrt{3} \times \sqrt{6} = 6$ .

Challenge

Reasoning

Harder reasoning

Expand and simplify $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})$ . What does the pattern resemble?
Expand $(2 + \sqrt{5})^2$ .
Rationalise $\dfrac{1}{2 + \sqrt{3}}$ . (Hint: multiply top and bottom by $2 - \sqrt{3}$ .)
The hypotenuse of a right-angled triangle with legs $1$ and $\sqrt{3}$ equals the side of a square. Find the area of the square.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Simplify

$2\sqrt{3}$ .
$3\sqrt{3}$ .
$3\sqrt{5}$ .
$7\sqrt{2}$ .
$2\sqrt{32} = 2 \times 4\sqrt{2} = 8\sqrt{2}$ .
$10\sqrt{2}$ .

Add & subtract

$4\sqrt{5}$ .
$3\sqrt{2}$ .
$\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$ .
$5\sqrt{2} - 4\sqrt{2} = \sqrt{2}$ .
$5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$ .

Multiply

$\sqrt{21}$ .
$6\sqrt{10}$ .
$6$ .
$\sqrt{16} = 4$ .
$2\sqrt{3} + \sqrt{36} = 2\sqrt{3} + 6$ .

Rationalise

$\dfrac{\sqrt{3}}{3}$ .
$\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$ .
$\dfrac{\sqrt{2}}{\sqrt{8}} = \sqrt{\tfrac{2}{8}} = \sqrt{\tfrac{1}{4}} = \tfrac{1}{2}$ .

Mixed & apply

side $= \sqrt{48} = 4\sqrt{3}$ cm.
diagonal $= \sqrt{9 + 25} = \sqrt{34}$ . Already in simplest form.
$\sqrt{18} + \sqrt{8} - \sqrt{50} = 3\sqrt{2} + 2\sqrt{2} - 5\sqrt{2} = 0$ .
$\sqrt{2} \times \sqrt{3} \times \sqrt{6} = \sqrt{2 \times 3 \times 6} = \sqrt{36} = 6$ .

Challenge

$(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$ . It’s the difference of squares pattern $(a+b)(a-b) = a^2 - b^2$ , used for rationalising binomial surds.
$(2 + \sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}$ .
$\dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$ .
Hypotenuse $= \sqrt{1 + 3} = 2$ . Area of square $= 2^2 = 4$ .

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