1MA1 · N8NumberHigherGrades 6–9

Surds

Your calculator says $\sqrt{2} = 1.414213562\dots$ , but that decimal never ends, so any rounded version is slightly wrong. A $\textbf{surd}$ like $\sqrt{2}$ keeps the value exact. Higher-tier questions often ask for answers "in the form $a + b\sqrt{c}$ ", which means working confidently with surds by hand.

50 min Powers, Roots & Standard Form

The big picture

A surd is an irrational root — one that cannot be written as a fraction. Surds behave like algebra: $3\sqrt5 + 2\sqrt5 = 5\sqrt5$ in the same way that $3x + 2x = 5x$ , and expanding $(2 + \sqrt3)(1 - \sqrt3)$ works like expanding double brackets.

The two key techniques are simplifying (pulling square factors out of the root) and rationalising (removing roots from a denominator). Both appear in exact-answer questions on Pythagoras, trigonometry and area, not just in surd questions.

What you'll learn

Simplify surds using $\sqrt{ab} = \sqrt{a}\sqrt{b}$
Add, subtract, multiply and divide surds
Expand brackets involving surds
Rationalise denominators of the form $\sqrt{a}$ and $a \pm \sqrt{b}$

Simplifying surds

Since $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ , look for the largest square number that is a factor. $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ .

A surd is fully simplified when the number under the root has no square factor other than 1.

Also useful: $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ , so $\sqrt{\dfrac{9}{16}} = \dfrac34$ .

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

Split a root over a product.

\sqrt{a} \times \sqrt{a} = a

A surd squared.

Example 1Simplify

\sqrt{200}

1The largest square factor of 200 is 100.

\sqrt{100 \times 2} = 10\sqrt2

Answer

10\sqrt2

Your turnSame method, new numbers — try it before moving on.

Simplify $\sqrt{98}$ .

Your answer

Type √ as sqrt, e.g. 5sqrt3.

Show worked solution

Solution

1 $98 = 49 \times 2$ .
2 $\sqrt{98} = 7\sqrt{2}$ .

Answer: $7\sqrt{2}$

Tip — If you pick a smaller square factor, you will need to simplify again: $\sqrt{200} = 2\sqrt{50} = 2 \times 5\sqrt2$ . Look for the largest one first.

Adding, subtracting and multiplying

Only like surds can be added or subtracted — simplify first to find them. $\sqrt{50} + \sqrt{18} = 5\sqrt2 + 3\sqrt2 = 8\sqrt2$ .

Multiply whole numbers with whole numbers and roots with roots: $3\sqrt2 \times 4\sqrt5 = 12\sqrt{10}$ .

Expand brackets exactly as in algebra, using $\sqrt{a} \times \sqrt{a} = a$ to simplify.

Example 1Expand and simplify

(3 + \sqrt5)(2 - \sqrt5)

3 \times 2 = 6

;

3 \times (-\sqrt5) = -3\sqrt5

;

\sqrt5 \times 2 = 2\sqrt5

;

\sqrt5 \times (-\sqrt5) = -5

2Collect:

6 - 5 - 3\sqrt5 + 2\sqrt5

Answer

1 - \sqrt5

Your turnSame method, new numbers — try it before moving on.

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

Your answer

Type it like 7 + 3sqrt5.

Show worked solution

Solution

1 $8 + 4\sqrt{3} - 2\sqrt{3} - 3$ .
2 $= 5 + 2\sqrt{3}$ .

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

Why this works

$\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ , but $\sqrt{9 + 16} = \sqrt{25} = 5$ . Roots do not split over addition — only over multiplication.

Rationalising a single surd

"Rationalise the denominator" means rewrite the fraction with no surd on the bottom.

For $\dfrac{a}{\sqrt{b}}$ , multiply top and bottom by $\sqrt{b}$ : the denominator becomes $b$ .

Example 1Rationalise

\dfrac{12}{\sqrt3}

1Multiply by

\dfrac{\sqrt3}{\sqrt3}

\dfrac{12\sqrt3}{3}

2Simplify.

Answer

4\sqrt3

Your turnSame method, new numbers — try it before moving on.

Rationalise the denominator of $\dfrac{15}{\sqrt{5}}$ .

Your answer

Type √ as sqrt, e.g. 2sqrt3.

Show worked solution

Solution

1 $\dfrac{15\sqrt{5}}{5}$ .
2 $= 3\sqrt{5}$ .

Answer: $3\sqrt{5}$

Rationalising with brackets

For a denominator like $a + \sqrt{b}$ , multiply top and bottom by the $\textbf{conjugate}$ $a - \sqrt{b}$ . The denominator becomes a difference of two squares, $a^{2} - b$ , with no surd.

Expand the numerator carefully and simplify fully at the end.

(a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b

Difference of two squares removes the surd.

Example 1Write

\dfrac{4}{3 - \sqrt5}

in the form

a + b\sqrt5

1Multiply top and bottom by

3 + \sqrt5

2Denominator:

9 - 5 = 4

3Numerator:

4(3 + \sqrt5) = 12 + 4\sqrt5

4Divide by 4.

Answer

3 + \sqrt5

Your turnSame method, new numbers — try it before moving on.

Write $\dfrac{6}{3 + \sqrt{3}}$ in the form $a + b\sqrt{3}$ .

Your answer

Type it like 3 - sqrt3.

Show hint

Multiply top and bottom by $3 - \sqrt{3}$ .

Show worked solution

Solution

1Denominator: $9 - 3 = 6$ .
2Numerator: $6(3 - \sqrt{3})$ . Divide by 6: $3 - \sqrt{3}$ .

Answer: $3 - \sqrt{3}$

Tip — Change only the sign between the terms to form the conjugate. $3 - \sqrt5$ becomes $3 + \sqrt5$ , not $-3 + \sqrt5$ .

Think like an examiner

Show the square factor used when simplifying, e.g. $\sqrt{72} = \sqrt{36}\sqrt{2}$ .
Write the multiplication by the conjugate explicitly as a fraction equal to 1.
Give answers in exactly the form requested, with integers $a$ and $b$ .

Surd rules

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

Products.

\tfrac{1}{\sqrt{a}} = \tfrac{\sqrt{a}}{a}

Rationalising.

(a + \sqrt{b})(a - \sqrt{b}) = a^{2} - b

Conjugates.

Watch out for these

Writing

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

Roots do not add like this; unlike surds cannot be combined.

Simplifying

\sqrt{48}

2\sqrt{12}

and stopping.

Use the largest square factor:

4\sqrt3

Writing

(\sqrt{7})^{2} = 49

(\sqrt{7})^{2} = 7

Multiplying only the denominator by the conjugate.

Multiply top and bottom so the value is unchanged.

Stretch yourself

A rectangle has sides $(5 + \sqrt3)$ cm and $(5 - \sqrt3)$ cm. Find, in simplest surd form, (a) its area and (b) the length of its diagonal.

Hint — The area is a difference of two squares. For the diagonal, use Pythagoras and expand each square.

Reveal solution

Solution(a) Area

= (5 + \sqrt3)(5 - \sqrt3) = 25 - 3 = 22

cm². (b)

d^{2} = (5 + \sqrt3)^{2} + (5 - \sqrt3)^{2} = (25 + 10\sqrt3 + 3) + (25 - 10\sqrt3 + 3) = 56

. So

d = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}

cm. Notice the surd parts cancel in both answers.

\blacksquare

Questions students ask

Why do we rationalise denominators?

It gives a standard form for exact answers, making them easier to compare, and historically made hand calculation easier.

\sqrt{16}

a surd?

No. It equals 4 exactly. A surd is a root that is irrational.

Can a surd answer be left unsimplified?

Usually no — "simplest form" means the number under the root has no square factor.

Key takeaways

Surds are irrational roots, used to keep answers exact.
Simplify by extracting the largest square factor.
Add or subtract only like surds; multiply roots with roots.
Rationalise $\tfrac{a}{\sqrt{b}}$ by multiplying by $\tfrac{\sqrt{b}}{\sqrt{b}}$ .
Rationalise $a \pm \sqrt{b}$ with the conjugate, using the difference of two squares.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 62 marks

Write $\sqrt{45} + \sqrt{20}$ in the form $a\sqrt{5}$ .[2]
Show mark scheme
- M1 $3\sqrt{5}$ or $2\sqrt{5}$
- A1 $5\sqrt{5}$
Answer: $5\sqrt{5}$

Question 2Grade 72 marks

Expand and simplify $(2\sqrt{3} - 1)^2$ .[2]
Show mark scheme
- M1Four terms with at least three correct: $12 - 2\sqrt{3} - 2\sqrt{3} + 1$
- A1 $13 - 4\sqrt{3}$
Answer: $13 - 4\sqrt{3}$

Question 3Grade 93 marks

Show that $\dfrac{1 + \sqrt{2}}{\sqrt{2} - 1}$ can be written as $3 + 2\sqrt{2}$ .[3]
Show mark scheme
- M1Multiplies by $\frac{\sqrt{2} + 1}{\sqrt{2} + 1}$
- M1Denominator $= 1$ ; numerator expanded $\sqrt{2} + 1 + 2 + \sqrt{2}$
- A1 $3 + 2\sqrt{2}$ shown

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