Maths · 1 Number

1.17 Surds

Before you start

The square numbers up to $12^{2} = 144$ , and their square roots.
Rational and irrational numbers: a rational number can be written as a fraction $\frac{p}{q}$ of two integers.
Expanding two brackets and collecting like terms, e.g. $(x + 3) (x - 2) = x^{2} + x - 6$ .
Multiplying the top and the bottom of a fraction by the same number gives an equal fraction, e.g. $\frac{3}{4} = \frac{6}{8}$ .
Pythagoras' theorem for a right-angled triangle with hypotenuse $c$ : $a^{2} + b^{2} = c^{2}$ .

Syllabus objectives covered (2)

Understand and use surds, including simplifying expressions.
Rationalise the denominator.

Part 1 of 7

What a surd is, and why you keep it exact

Some square roots are whole numbers: $\sqrt{9} = 3$ and $\sqrt{49} = 7$ . Others are not. $\sqrt{2}$ is about $1.414$ , and its decimal never ends and never repeats. A number like that is irrational: it cannot be written as a fraction of two integers.
A surd is a root that is irrational. $\sqrt{2}$ , $\sqrt{5}$ and $\sqrt{12}$ are surds. $\sqrt{16} = 4$ and $\sqrt{0.25} = 0.5$ are not surds, because they are rational. The surds in this lesson are all square roots.
The symbol $\sqrt{a}$ means the positive square root of $a$ . So $\sqrt{25}$ is $5$ , not $- 5$ .
Why keep $\sqrt{2}$ as it is? Because any decimal you write for it is only an approximation. ${1.414}^{2} = 1.999396$ , which is not $2$ . The surd is the exact value: $\sqrt{2} \times \sqrt{2} = 2$ exactly.
When a question asks for an exact answer, or for an answer in surd form, it wants the surd, not a decimal. A decimal is not in the form the question asks for.
Treat a surd like a letter in algebra. $3 \sqrt{2}$ means $3 \times \sqrt{2}$ , in the same way that $3 x$ means $3 \times x$ .

Part 2 of 7

Multiplying and dividing surds

Two square roots multiply into one square root. This is true for any positive numbers $a$ and $b$ :
$\sqrt{a} \times \sqrt{b} = \sqrt{a b}$
Here is why. Square the left-hand side: ${(\sqrt{a} \times \sqrt{b})}^{2} = \sqrt{a} \sqrt{a} \times \sqrt{b} \sqrt{b} = a \times b$ . So $\sqrt{a} \times \sqrt{b}$ is a positive number whose square is $a b$ . That is exactly what $\sqrt{a b}$ means.
Two surds can multiply to give a whole number: $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$ .
The special case you will use most: a square root times itself is the number under the root.
$\sqrt{a} \times \sqrt{a} = a$
Division works the same way. For example, $\frac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5$ .
$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
When there are whole numbers in front, multiply the whole numbers together and the surds together: $2 \sqrt{3} \times 4 \sqrt{3} = 8 \times 3 = 24$ . For division, divide the whole numbers and divide the surds: $\frac{6 \sqrt{10}}{2 \sqrt{5}} = 3 \sqrt{2}$ .
The trap: addition does not work like this. $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ , but $\sqrt{9 + 16} = \sqrt{25} = 5$ . So $\sqrt{a} + \sqrt{b}$ is not $\sqrt{a + b}$ . Only multiplication and division join two roots into one.

Part 3 of 7

Simplifying a surd

A surd is in its simplest form when the number under the root has no square factor except 1. $\sqrt{20}$ is not in its simplest form, because $20 = 4 \times 5$ and $4$ is a square number.
To simplify, split the number into a square number times another number. Then use $\sqrt{a b} = \sqrt{a} \times \sqrt{b}$ backwards, and take the square root of the square number.
$\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2 \sqrt{5}$
Look for the largest square factor. Test the square numbers $4, 9, 16, 25, 36, 49, 64, 81, 100$ , starting from the largest one that could divide the number.
If you use a smaller square factor first, you are not finished. $\sqrt{72} = \sqrt{4} \times \sqrt{18} = 2 \sqrt{18}$ , but $18 = 9 \times 2$ , so $2 \sqrt{18} = 2 \times 3 \sqrt{2} = 6 \sqrt{2}$ . Using $36$ at the start gives $\sqrt{36} \times \sqrt{2} = 6 \sqrt{2}$ in one step.
'Simplify' means simplify fully. $2 \sqrt{18}$ has the right value, but it is not simplified, so it does not answer the question.
Simplify the result of a multiplication too: $\sqrt{3} \times \sqrt{6} = \sqrt{18} = \sqrt{9} \times \sqrt{2} = 3 \sqrt{2}$ .
You can also go backwards: move a whole number inside the root by squaring it. $3 \sqrt{2} = \sqrt{9} \times \sqrt{2} = \sqrt{18}$ .
This is how you compare two surds without a calculator. $3 \sqrt{2} = \sqrt{18}$ and $2 \sqrt{5} = \sqrt{20}$ . Since $20 > 18$ , $2 \sqrt{5}$ is the larger.

Part 4 of 7

Adding and subtracting surds

You can add or subtract surds only when they have the same number under the root. These are like surds, and you collect them as you collect like terms in algebra, in the same way that $5 x + 2 x = 7 x$ .
$5 \sqrt{3} + 2 \sqrt{3} = 7 \sqrt{3}$
Unlike surds cannot be collected. $3 \sqrt{5} + 2 \sqrt{3}$ is already as simple as it can be, just as $3 x + 2 y$ is.
Surds that look unlike can become like surds once you simplify them. So simplify every surd first, then collect.
$\sqrt{200} - \sqrt{32} = 10 \sqrt{2} - 4 \sqrt{2} = 6 \sqrt{2}$
Here $200 = 100 \times 2$ and $32 = 16 \times 2$ . Both simplify to a multiple of $\sqrt{2}$ , so they can be collected.
A whole number and a surd are unlike terms too. $4 + 3 \sqrt{2}$ does not simplify to $7 \sqrt{2}$ .

Part 5 of 7

Expanding brackets that contain surds

Expand surd brackets exactly as you expand $(x + 3) (x - 2)$ : multiply every term in the first bracket by every term in the second. Then replace each $\sqrt{a} \times \sqrt{a}$ with $a$ and collect like terms.
$(3 + \sqrt{2}) (1 + \sqrt{2}) = 3 + 3 \sqrt{2} + \sqrt{2} + 2 = 5 + 4 \sqrt{2}$
The last product is $\sqrt{2} \times \sqrt{2} = 2$ . It is a whole number, so it collects with the $3$ , not with the surds.
Squaring a bracket means multiplying it by itself. Write it as two brackets, so that you cannot forget the middle terms.
${(\sqrt{5} + 1)}^{2} = (\sqrt{5} + 1) (\sqrt{5} + 1) = 5 + \sqrt{5} + \sqrt{5} + 1 = 6 + 2 \sqrt{5}$
The trap is to square each term and write $5 + 1 = 6$ . That leaves out the two middle terms, which add up to $2 \sqrt{5}$ .
Two brackets that differ only in the sign in the middle are special. The middle terms cancel, and no surd is left.
$(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - a \sqrt{b} + a \sqrt{b} - b = a^{2} - b$
For example, $(3 + \sqrt{2}) (3 - \sqrt{2}) = 9 - 2 = 7$ . This is the difference of two squares, and it is the key to rationalising a denominator with two terms.

Part 6 of 7

Rationalising a denominator that is a single surd

To rationalise the denominator of a fraction is to rewrite the fraction so that no surd is left on the bottom. The value of the fraction does not change, only the way it is written.
If the denominator is $\sqrt{b}$ , multiply the top and the bottom by $\sqrt{b}$ . The bottom becomes $\sqrt{b} \times \sqrt{b} = b$ , a whole number.
This is allowed because $\frac{\sqrt{b}}{\sqrt{b}} = 1$ , and multiplying by 1 does not change a number. It is the same move as writing $\frac{3}{4} = \frac{6}{8}$ .
$\frac{10}{\sqrt{5}} = \frac{10 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{10 \sqrt{5}}{5} = 2 \sqrt{5}$
After you rationalise, look for a cancellation. Here $10 \div 5 = 2$ .
If a whole number is in front of the surd on the bottom, multiply by the surd only. For $\frac{3}{2 \sqrt{6}}$ , multiply the top and the bottom by $\sqrt{6}$ . The bottom becomes $2 \times 6 = 12$ , so $\frac{3}{2 \sqrt{6}} = \frac{3 \sqrt{6}}{12} = \frac{\sqrt{6}}{4}$ .
The same idea gives the exact value of the square root of a fraction: $\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .
Multiply the top and the bottom. Multiplying only the bottom changes the value of the fraction, so the answer is wrong.

Part 7 of 7

Rationalising a denominator with two terms

Multiplying $\frac{1}{3 + \sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ does not help. The bottom becomes $3 \sqrt{2} + 2$ , which still contains a surd.
Use the difference of two squares instead. Multiply the top and the bottom by the same two terms with the sign in the middle changed. The pair $3 + \sqrt{2}$ and $3 - \sqrt{2}$ are called conjugates.
$\frac{1}{3 + \sqrt{2}} = \frac{1 \times (3 - \sqrt{2})}{(3 + \sqrt{2}) (3 - \sqrt{2})} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$
The bottom is always $a^{2} - b$ , a whole number, because the two middle terms cancel. You can write it straight down as $a^{2} - b$ , but do write it down.
Example: $\frac{1}{- 1 + \sqrt{3}}$ . Write the denominator as $\sqrt{3} - 1$ , so the conjugate is $\sqrt{3} + 1$ .
$\frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{1 + \sqrt{3}}{2}$
Multiplying by $- 1 - \sqrt{3}$ also works. The bottom becomes ${(- 1)}^{2} - 3 = - 2$ and the top becomes $- 1 - \sqrt{3}$ , which gives the same answer, $\frac{1 + \sqrt{3}}{2}$ . What matters is that the sign in front of the surd changes.
The denominator can have two surds, as in $\sqrt{5} + \sqrt{2}$ . Change the sign in the middle in the same way: $(\sqrt{5} + \sqrt{2}) (\sqrt{5} - \sqrt{2}) = 5 - 2 = 3$ .
If the top also has two terms, put it in brackets and expand it with the method from the brackets section. Do not try to cancel a term on the top with a term on the bottom.
A question can ask for the answer in a form such as $a + b \sqrt{5}$ . Divide every term on the top by the bottom, then read off the values. For example, $\frac{8 + 4 \sqrt{5}}{4} = 2 + \sqrt{5}$ , so $a = 2$ and $b = 1$ .

Worked examples

Example 1. Simplify (a) $\sqrt{75}$ (b) $\sqrt{48} - \sqrt{27} + \sqrt{75}$ (c) $3 \sqrt{2} \times 5 \sqrt{6}$

Show the working

(a) Find the largest square number that divides 75. That is 25, because $75 = 25 \times 3$ .
$\sqrt{75} = \sqrt{25} \times \sqrt{3} = 5 \sqrt{3}$
The number left under the root, 3, has no square factor, so this is the simplest form.
(b) Simplify each surd first. $48 = 16 \times 3$ and $27 = 9 \times 3$ , and $\sqrt{75} = 5 \sqrt{3}$ from part (a).
$\sqrt{48} - \sqrt{27} + \sqrt{75} = 4 \sqrt{3} - 3 \sqrt{3} + 5 \sqrt{3}$
All three are now multiples of $\sqrt{3}$ , so they are like surds. Collect them: $4 - 3 + 5 = 6$ .
$= 6 \sqrt{3}$
(c) Multiply the whole numbers together and the surds together.
$3 \sqrt{2} \times 5 \sqrt{6} = 15 \sqrt{12}$
This is not finished, because $12 = 4 \times 3$ . So $\sqrt{12} = 2 \sqrt{3}$ .
$15 \sqrt{12} = 15 \times 2 \sqrt{3} = 30 \sqrt{3}$

Answer: (a) $5 \sqrt{3}$ (b) $6 \sqrt{3}$ (c) $30 \sqrt{3}$

Example 2. Expand and simplify (a) $(2 + \sqrt{3}) (5 - \sqrt{3})$ (b) ${(\sqrt{7} - 2)}^{2}$

Show the working

(a) Multiply each term in the first bracket by each term in the second. There are four products.
$2 \times 5 = 10, 2 \times (- \sqrt{3}) = - 2 \sqrt{3}, \sqrt{3} \times 5 = 5 \sqrt{3}, \sqrt{3} \times (- \sqrt{3}) = - 3$
The last product is $- 3$ because $\sqrt{3} \times \sqrt{3} = 3$ and the sign is negative.
Collect the whole numbers and the surds separately.
$10 - 3 - 2 \sqrt{3} + 5 \sqrt{3} = 7 + 3 \sqrt{3}$
(b) Write the square as two brackets.
$(\sqrt{7} - 2) (\sqrt{7} - 2) = 7 - 2 \sqrt{7} - 2 \sqrt{7} + 4$
The last product is $(- 2) \times (- 2) = + 4$ . Collect the terms.
$= 11 - 4 \sqrt{7}$
Squaring each term on its own would give $7 + 4 = 11$ and miss the middle terms. The full answer has $- 4 \sqrt{7}$ as well.

Answer: (a) $7 + 3 \sqrt{3}$ (b) $11 - 4 \sqrt{7}$

Example 3. Rationalise the denominator and simplify. (a) $\frac{12}{\sqrt{6}}$ (b) $\frac{5}{3 \sqrt{10}}$ (c) $\frac{6}{\sqrt{5} + \sqrt{2}}$

Show the working

(a) Multiply the top and the bottom by $\sqrt{6}$ . The bottom becomes 6.
$\frac{12 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{12 \sqrt{6}}{6} = 2 \sqrt{6}$
(b) Multiply the top and the bottom by $\sqrt{10}$ only. The 3 is already a whole number. The bottom becomes $3 \times 10 = 30$ .
$\frac{5 \times \sqrt{10}}{3 \sqrt{10} \times \sqrt{10}} = \frac{5 \sqrt{10}}{30}$
Cancel the common factor 5 from 5 and 30.
$\frac{5 \sqrt{10}}{30} = \frac{\sqrt{10}}{6}$
(c) The bottom has two terms, so multiply the top and the bottom by $\sqrt{5} - \sqrt{2}$ , the same terms with the sign in the middle changed.
$\frac{6 (\sqrt{5} - \sqrt{2})}{(\sqrt{5} + \sqrt{2}) (\sqrt{5} - \sqrt{2})} = \frac{6 (\sqrt{5} - \sqrt{2})}{5 - 2}$
The bottom is $5 - 2 = 3$ . Divide the 6 by 3.
$\frac{6 (\sqrt{5} - \sqrt{2})}{3} = 2 (\sqrt{5} - \sqrt{2}) = 2 \sqrt{5} - 2 \sqrt{2}$

Answer: (a) $2 \sqrt{6}$ (b) $\frac{\sqrt{10}}{6}$ (c) $2 \sqrt{5} - 2 \sqrt{2}$

Example 4. A rectangle has an area of 6 cm² and a width of $(3 - \sqrt{3})$ cm. (a) Find the length of the rectangle. Give your answer in the form $a + b \sqrt{3}$ . (b) Find the perimeter of the rectangle. (c) Find the length of a diagonal of the rectangle. Give your answer as a surd in its simplest form.

Show the working

(a) The area is length × width, so the length is the area divided by the width.
$length = \frac{6}{3 - \sqrt{3}}$
Rationalise: multiply the top and the bottom by $3 + \sqrt{3}$ . The bottom is $3^{2} - 3 = 6$ .
$\frac{6 (3 + \sqrt{3})}{(3 - \sqrt{3}) (3 + \sqrt{3})} = \frac{6 (3 + \sqrt{3})}{6} = 3 + \sqrt{3}$
Check: $(3 + \sqrt{3}) (3 - \sqrt{3}) = 9 - 3 = 6$ , which is the area. Here $a = 3$ and $b = 1$ .
(b) The perimeter is 2 × (length + width).
$2 ((3 + \sqrt{3}) + (3 - \sqrt{3})) = 2 \times 6 = 12$
The surds cancel, so the perimeter is a whole number of centimetres.
(c) A diagonal is the hypotenuse of a right-angled triangle whose shorter sides are the length and the width. Use Pythagoras.
$d^{2} = {(3 + \sqrt{3})}^{2} + {(3 - \sqrt{3})}^{2}$
Expand each square as two brackets: ${(3 + \sqrt{3})}^{2} = 9 + 6 \sqrt{3} + 3 = 12 + 6 \sqrt{3}$ , and ${(3 - \sqrt{3})}^{2} = 12 - 6 \sqrt{3}$ .
$d^{2} = 12 + 6 \sqrt{3} + 12 - 6 \sqrt{3} = 24$
So $d = \sqrt{24}$ . The largest square factor of 24 is 4.
$d = \sqrt{4} \times \sqrt{6} = 2 \sqrt{6}$

Answer: (a) $(3 + \sqrt{3})$ cm (b) 12 cm (c) $2 \sqrt{6}$ cm

Where people go wrong

Mistake: Adding the numbers under the roots, e.g. writing $\sqrt{9} + \sqrt{16} = \sqrt{25}$ .

Instead: Work out roots that are whole numbers: $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ . Add surds only when they are like surds, as with algebra terms: $5 \sqrt{3} + 2 \sqrt{3} = 7 \sqrt{3}$ .

The rule $\sqrt{a} \times \sqrt{b} = \sqrt{a b}$ is for multiplication only. $\sqrt{25} = 5$ , not $7$ , so an addition rule like it is false.

Mistake: Squaring a bracket term by term, e.g. ${(\sqrt{5} + 1)}^{2} = 5 + 1 = 6$ .

Instead: Write it as $(\sqrt{5} + 1) (\sqrt{5} + 1)$ and find all four products: $5 + \sqrt{5} + \sqrt{5} + 1 = 6 + 2 \sqrt{5}$ .

Squaring a bracket multiplies every term by every term. The two middle terms are part of the answer.

Mistake: Multiplying only the bottom by the surd, e.g. $\frac{10}{\sqrt{5}} = \frac{10}{5} = 2$ .

Instead: Multiply the top and the bottom: $\frac{10 \sqrt{5}}{5} = 2 \sqrt{5}$ .

Multiplying the top and the bottom by the same number is multiplying by 1, which keeps the value. Changing the bottom alone gives a different number: $\frac{10}{\sqrt{5}}$ is about $4.47$ , not $2$ .

Mistake: Multiplying by the denominator itself, e.g. $\frac{1}{3 + \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}$ .

Instead: Change the sign in the middle: multiply by $\frac{3 - \sqrt{2}}{3 - \sqrt{2}}$ , which makes the bottom $9 - 2 = 7$ .

${(3 + \sqrt{2})}^{2} = 11 + 6 \sqrt{2}$ , which still contains a surd. Only the changed sign makes the middle terms cancel.

Mistake: Stopping before a surd is fully simplified, e.g. $\sqrt{72} = 2 \sqrt{18}$ .

Instead: Test the number left under the root for a square factor again: $18 = 9 \times 2$ , so $2 \sqrt{18} = 6 \sqrt{2}$ . Or use the largest square factor, 36, at the start.

'Simplify' means simplify fully. A surd is in its simplest form only when the number under the root has no square factor except 1.

Check yourself

Which of these numbers are surds? $\sqrt{16}$ , $\sqrt{15}$ , $\sqrt{0.25}$ , $\sqrt{50}$ , $\sqrt{49}$
Answer
$\sqrt{15}$ and $\sqrt{50}$ . The others are rational: $\sqrt{16} = 4$ , $\sqrt{0.25} = 0.5$ and $\sqrt{49} = 7$ .
Simplify $\sqrt{98}$ .
Answer
$7 \sqrt{2}$ , because $98 = 49 \times 2$ .
Simplify $\sqrt{45} + \sqrt{20}$ .
Answer
$5 \sqrt{5}$ . $\sqrt{45} = 3 \sqrt{5}$ and $\sqrt{20} = 2 \sqrt{5}$ , and $3 + 2 = 5$ .
Simplify (a) $\sqrt{6} \times \sqrt{15}$ (b) $\frac{\sqrt{60}}{\sqrt{5}}$
Answer
(a) $3 \sqrt{10}$ , because $\sqrt{6} \times \sqrt{15} = \sqrt{90} = \sqrt{9} \times \sqrt{10}$ . (b) $2 \sqrt{3}$ , because $\frac{\sqrt{60}}{\sqrt{5}} = \sqrt{12} = \sqrt{4} \times \sqrt{3}$ .
Expand and simplify $(4 + \sqrt{2}) (4 - \sqrt{2})$ .
Answer
$14$ . The middle terms $- 4 \sqrt{2}$ and $+ 4 \sqrt{2}$ cancel, which leaves $16 - 2$ .
Rationalise the denominator of $\frac{9}{\sqrt{3}}$ and simplify.
Answer
$3 \sqrt{3}$ . Multiplying the top and the bottom by $\sqrt{3}$ gives $\frac{9 \sqrt{3}}{3} = 3 \sqrt{3}$ .
Rationalise the denominator of $\frac{2}{\sqrt{5} + 1}$ and simplify.
Answer
$\frac{\sqrt{5} - 1}{2}$ . Multiply the top and the bottom by $\sqrt{5} - 1$ . The bottom is $5 - 1 = 4$ , so you get $\frac{2 (\sqrt{5} - 1)}{4} = \frac{\sqrt{5} - 1}{2}$ .
Without a calculator, explain which is larger, $3 \sqrt{5}$ or $5 \sqrt{2}$ .
Answer
$5 \sqrt{2}$ . Move each whole number inside the root by squaring it: $3 \sqrt{5} = \sqrt{45}$ and $5 \sqrt{2} = \sqrt{50}$ . Since $50 > 45$ , $5 \sqrt{2}$ is larger.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Simplify $\sqrt{180}$ . 2 marks No calculator

Mark scheme

M1 √36 × √5 or √(36 × 5) seen, or a correct partial simplification such as 2√45 or 3√20
A1 6√5 cao

(b) Simplify $\sqrt{128} - \sqrt{18}$ . 2 marks No calculator

Mark scheme

M1 8√2 or 3√2 seen
A1 5√2 cao

(c) Expand and simplify $(5 - 2 \sqrt{3}) (1 + \sqrt{3})$ . 2 marks No calculator

Mark scheme

M1 at least 3 of the 4 terms 5, 5√3, −2√3, −6 correct
A1 −1 + 3√3 oe, e.g. 3√3 − 1

Question 2.

(a) Write $\frac{14}{\sqrt{7}}$ in the form $k \sqrt{7}$ , where $k$ is an integer. 2 marks No calculator

Mark scheme

M1 numerator and denominator multiplied by √7, or 14√7/7 seen
A1 2√7 cao

(b) Write $\frac{11}{4 + \sqrt{5}}$ in the form $a + b \sqrt{5}$ , where $a$ and $b$ are integers. 3 marks No calculator

Mark scheme

M1 numerator and denominator multiplied by (4 − √5) oe
A1 denominator (4 + √5)(4 − √5) = 11 seen, or 16 − 5 seen
A1 4 − √5, or a = 4 and b = −1

(c) Show that $\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$ . 3 marks No calculator

Mark scheme

M1 numerator and denominator multiplied by (√3 + 1) oe; or, alternative method, (2 + √3)(√3 − 1) expanded
A1 numerator 4 + 2√3 and denominator 2 both seen; or, alternative method, 2√3 − 2 + 3 − √3 seen
A1 (4 + 2√3)/2 = 2 + √3 completed with no errors seen; or, alternative method, = √3 + 1 so the result follows, with no errors seen

Question 3. A right-angled triangle has two shorter sides of length $(\sqrt{5} + 1)$ cm and $(\sqrt{5} - 1)$ cm.

(a) Work out the area of the triangle. 2 marks No calculator

Mark scheme

M1 ½ × (√5 + 1)(√5 − 1) oe, or (√5 + 1)(√5 − 1) = 4 seen
A1 2 (cm²) cao

(b) Find the length of the hypotenuse of the triangle. Give your answer as a surd in its simplest form. 3 marks No calculator

Mark scheme

M1 (√5 + 1)² + (√5 − 1)² oe
B1 6 + 2√5 or 6 − 2√5 seen
A1 2√3 (cm) cao; √12 is not in its simplest form and does not earn this mark

How this is examined

Rationalising a two-term denominator by multiplying by the same terms with the sign changed: A Write part of about 3 marks that asks for the answer in a given form, e.g. 'in the form $a + b \sqrt{5}$ , where $a$ and $b$ are integers' The multiplication written on the top and the bottom, e.g. $\frac{11 (4 - \sqrt{5})}{(4 + \sqrt{5}) (4 - \sqrt{5})}$ ; the bottom worked out as a whole number, $16 - 5 = 11$ ; then the answer in exactly the form asked for, $4 - \sqrt{5}$ . A decimal is not in the required form.
Showing a surd result that the question gives you: A Show (that) of 2 or 3 marks with the simplified result printed in the question Every step written, with no decimals: the multiplication by the conjugate, the expanded top, e.g. $4 + 2 \sqrt{3}$ , the whole-number bottom, e.g. $2$ , then a last line that matches the given result. The given result written without the steps that lead to it does not show it.
Exact lengths and areas when the sides of a shape are surd expressions: A Work out or Find part on a rectangle or a right-angled triangle, with the answer asked for as a surd in its simplest form Pythagoras or the area formula written with each surd expression in brackets, each square expanded correctly, e.g. ${(\sqrt{5} + 1)}^{2} = 6 + 2 \sqrt{5}$ , then the exact answer in its simplest form, e.g. $\sqrt{12} = 2 \sqrt{3}$ . A decimal, or $\sqrt{12}$ left unsimplified, is not the final answer.
Writing each surd as a multiple of the same root, then collecting: A 2-mark Simplify on a sum or difference of two or three unsimplified surds Each surd simplified with its largest square factor, e.g. $\sqrt{128} = 8 \sqrt{2}$ and $\sqrt{18} = 3 \sqrt{2}$ , then the like surds collected into one term, $5 \sqrt{2}$ . An answer that is not fully simplified, such as $2 \sqrt{18}$ , is not the final answer.
Expanding two brackets that contain surds: A 2-mark Expand and simplify All four products written, each $\sqrt{a} \times \sqrt{a}$ replaced by $a$ , then the whole numbers and the surds collected separately into one answer, e.g. $- 1 + 3 \sqrt{3}$ .
Removing a single surd from the bottom of a fraction: A 2-mark Write in the form $k \sqrt{n}$ The top and the bottom multiplied by the surd, e.g. $\frac{14 \sqrt{7}}{7}$ seen, then the cancellation to $2 \sqrt{7}$ .

Words to know

Surd: A root that is irrational, so it cannot be written as a fraction and its decimal never ends or repeats, e.g. $\sqrt{2}$ or $\sqrt{12}$ .
Irrational number: A number that cannot be written as a fraction $\frac{p}{q}$ of two integers, e.g. $\sqrt{3}$ or $π$ .
Exact value: A value written without rounding, e.g. $\sqrt{2}$ rather than $1.414$ .
Simplest form of a surd: $k \sqrt{n}$ , where $n$ has no square factor except 1, e.g. $6 \sqrt{2}$ rather than $\sqrt{72}$ or $2 \sqrt{18}$ .
Like surds: Multiples of the same root, e.g. $4 \sqrt{3}$ and $- 3 \sqrt{3}$ . Only like surds can be added or subtracted.
Rationalise the denominator: Rewrite a fraction so that no surd is left on the bottom, without changing its value.
Conjugate: The same two terms with the sign in the middle changed: the conjugate of $3 + \sqrt{2}$ is $3 - \sqrt{2}$ . Their product, $9 - 2 = 7$ , has no surd.
Difference of two squares: $(a + b) (a - b) = a^{2} - b^{2}$ . The two middle terms cancel.

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