A surd is an exact irrational root; simplify it by extracting the largest square factor, as in √(50)=5√(2). Combine only like surds, and rationalise a binomial denominator by multiplying numerator and denominator by its conjugate.

How many marks are 2.2 questions worth?

2.2 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

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2.2

Use and manipulate surds, including rationalising the denominator.

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Surds

Worked answers and methods for 2.2 on Edexcel A-level Maths 9MA0.

Explanation

A surd is an exact irrational root; simplify it by extracting the largest square factor, as in $\sqrt{50}=5\sqrt2$ . Combine only like surds, and rationalise a binomial denominator by multiplying numerator and denominator by its conjugate.
For example, $1/(3+\sqrt2)=(3-\sqrt2)/(9-2)=(3-\sqrt2)/7$ .
Do not split roots across addition: $\sqrt{a+b}$ is not generally $\sqrt a+\sqrt b$ ; after squaring an equation, check every candidate in the original.
Keep surd answers exact unless the question requests a decimal.
In equations, record the domain before squaring and substitute every candidate back into the original radical equation because squaring can introduce an extraneous solution.

Worked example

Rationalise and simplify $\dfrac{5}{2+\sqrt3}$ .

1.Multiply by the conjugate: $\dfrac{5}{2+\sqrt3}\times\dfrac{2-\sqrt3}{2-\sqrt3}=\dfrac{5(2-\sqrt3)}{4-3}=10-5\sqrt3$ .

Answer: $10-5\sqrt3$

Common mistakes

Don't split $\sqrt{a+b}$ into $\sqrt a+\sqrt b$ , which is not a valid surd law.
Don't multiply a binomial denominator by itself instead of by its conjugate, so the denominator remains irrational.

Exam tip

For “exact” answers, simplify every surd and rationalise the denominator without converting to decimals.

Worked practice

Tier 1 · Easy

Write

\sqrt{72}

in the form

a\sqrt2

, where

a

is an integer.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$6\sqrt2$	2
Notes
Use the square factor $36$ : $\sqrt{72}=\sqrt{36\times2}=6\sqrt2$ .

(2 marks)

Tier 2 · Standard

Show that

\dfrac{2+\sqrt3}{2-\sqrt3}-\dfrac{2-\sqrt3}{2+\sqrt3}=8\sqrt3

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$8\sqrt3$	4
Notes
Rationalising each fraction gives $\dfrac{(2+\sqrt3)^2}{4-3}-\dfrac{(2-\sqrt3)^2}{4-3}=(7+4\sqrt3)-(7-4\sqrt3)=8\sqrt3$ , as required.

(4 marks)

Tier 3 · Hard

Solve

\sqrt{x+6}-\sqrt{x-3}=1

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=19$	5
Notes
The domain requires $x\geq3$ . Rearrange to $\sqrt{x+6}=1+\sqrt{x-3}$ and square: $x+6=x-2+2\sqrt{x-3}$ . Thus $8=2\sqrt{x-3}$ , so $\sqrt{x-3}=4$ and $x=19$ . Substitution gives $5-4=1$ , so the solution is valid.

(5 marks)

Tier 1 · Easy

Simplify

2\sqrt{45}-\sqrt{20}+\sqrt5

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$5\sqrt5$	2
Notes
$\sqrt{45}=3\sqrt5$ and $\sqrt{20}=2\sqrt5$ . Hence $2\sqrt{45}-\sqrt{20}+\sqrt5=6\sqrt5-2\sqrt5+\sqrt5=5\sqrt5$ .

(2 marks)

Tier 2 · Standard

Rationalise the denominator of

\dfrac{4+\sqrt7}{\sqrt7-2}

, giving your answer in exact simplified form.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$5+2\sqrt7$	3
Notes
Multiply numerator and denominator by $\sqrt7+2$ . The denominator becomes $7-4=3$ , while the numerator is $(4+\sqrt7)(\sqrt7+2)=15+6\sqrt7$ . Dividing by $3$ gives $5+2\sqrt7$ .

(3 marks)

Tier 3 · Hard

Let

y=\sqrt{11+6\sqrt2}

. Express

y

in the form

a+b\sqrt2

, where

a

and

b

are positive integers. Hence express

1/y

with a rational denominator.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$y=3+\sqrt2$ $\dfrac1y=\dfrac{3-\sqrt2}{7}$	5
Notes
If $y=a+b\sqrt2$ , then $y^2=a^2+2b^2+2ab\sqrt2$ . Matching $11+6\sqrt2$ gives $a^2+2b^2=11$ and $2ab=6$ , so $ab=3$ and $(a,b)=(3,1)$ or $(1,3)$ ; only $(3,1)$ satisfies $a^2+2b^2=11$ , since $1+18=19\neq11$ . Thus $a=3$ , $b=1$ , and $(3+\sqrt2)^2=11+6\sqrt2$ , so the positive square root is $y=3+\sqrt2$ . Therefore $1/y=(3-\sqrt2)/(9-2)=(3-\sqrt2)/7$ .

(5 marks)

Tier 2 · Standard

The rational numbers

a

and

b

satisfy

(a+b\sqrt3)(2-\sqrt3)=7+\sqrt3

. Find

a

and

b

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$a=17$ , $b=9$	4
Notes
Expanding gives $(2a-3b)+(2b-a)\sqrt3=7+\sqrt3$ . Equating rational and irrational parts gives $2a-3b=7$ and $2b-a=1$ . The second equation gives $a=2b-1$ ; substituting into the first gives $b=9$ , and hence $a=17$ .

(4 marks)

Tier 3 · Hard

Let

u=2+\sqrt3

. Show that

u^{-1}=2-\sqrt3

and hence find the exact value of

u^3+u^{-3}

without expanding both cubes separately.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$u^{-1}=2-\sqrt3$ $u^3+u^{-3}=52$	5
Notes
Since $(2+\sqrt3)(2-\sqrt3)=1$ , it follows that $u^{-1}=2-\sqrt3$ . Therefore $u+u^{-1}=4$ . Using $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=u$ , $b=u^{-1}$ and $ab=1$ gives $u^3+u^{-3}=4^3-3(1)(4)=52$ .

(5 marks)

Tier 3 · Hard

Rationalise the denominator of

\dfrac1{\sqrt2+\sqrt3+\sqrt5}

, giving your answer in exact simplified form.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\dfrac{2\sqrt3+3\sqrt2-\sqrt{30}}{12}$	5
Notes
First multiply numerator and denominator by $\sqrt2+\sqrt3-\sqrt5$ . The denominator becomes $(\sqrt2+\sqrt3)^2-5=2\sqrt6$ . The expression is therefore $(\sqrt2+\sqrt3-\sqrt5)/(2\sqrt6)$ . Multiplying by $\sqrt6/\sqrt6$ gives $(2\sqrt3+3\sqrt2-\sqrt{30})/12$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Let

\alpha=\sqrt5+\sqrt2

and

\beta=\sqrt5-\sqrt2

. Find a monic polynomial

P(x)

with integer coefficients whose four roots are

\alpha

-\alpha

\beta

and

-\beta

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$P(x)=x^4-14x^2+9$	6
Notes
The squared positive roots are $\alpha^2=7+2\sqrt{10}$ and $\beta^2=7-2\sqrt{10}$ . Hence $\alpha^2+\beta^2=14$ and $\alpha^2\beta^2=(7+2\sqrt{10})(7-2\sqrt{10})=9$ . Therefore $(x^2-\alpha^2)(x^2-\beta^2)=x^4-(\alpha^2+\beta^2)x^2+\alpha^2\beta^2=x^4-14x^2+9$ . Its roots are exactly the four stated values.

(6 marks)

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