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com

1. Write

√(75) – √(27)

in the form k √x, where k and x are integers.

(Total 2 marks)

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2. (a) Expand and simplify (7 + √5)(3 – √5).

(3)

7+ 5
(b) Express in the form a + b √5, where a and b are integers.
3+ 5
(3)
(Total 6 marks)

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3. Simplify

(a) (3 √7)2
(1)

(b) (8 + √5)(2 – √5)

(3)
(Total 4 marks)

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4. Given that 32 √2 = 2a, find the value of a.

(Total 3 marks)

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Algebra: Surds and Indices – Questions 2

 PhysicsAndMathsTutor.com

(3 – 4 x ) 2
5. f( x) = , x>0
x

1 1
–
(a) Show that f( x) = 9 x 2 + Ax 2 + B, where A and B are constants to be found.
(3)

(b) Find f´(x).

(3)

(c) Evaluate f´(9).

(2)
(Total 8 marks)

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1
6. (a) Write down the value of 125 3 .
(1)
2
−
(b) Find the value of 125 3 .
(2)
(Total 3 marks)

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7. Expand and simplify ( 7 + 2)( 7 − 2).

(Total 2 marks)

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Algebra: Surds and Indices – Questions 3

 PhysicsAndMathsTutor.com

3
2x 2 − x
8. Given that 2 can be written in the form 2 x p − x q ,
x

(a) write down the value of p and the value of q.

(2)
3
2x 2 − x 2
Given that y = 5 x 2 − 3 + ,
x

dy
(b) find , simplifying the coefficient of each term.
dx
(4)
(Total 6 marks)

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1
9. (a) Write down the value of 16 4 .
(1)

3
(b) Simplify (16 x12 ) 4 .
(2)
(Total 3 marks)

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10. Simplify

5− 3
,
2+ 3

giving your answer in the form a+b√3, where a and b are integers.
(Total 4 marks)

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11. Simplify (3 + √5)(3 – √5).

(Total 2 marks)

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Algebra: Surds and Indices – Questions 4

 PhysicsAndMathsTutor.com
4
12. (a) Find the value of 83 .
(2)
4
15 x 3
(b) Simplify .
3x
(2)
(Total 4 marks)

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13. (a) Express √108 in the form a√3, where a is an integer.

(1)

(b) Express (2 – √3)2 in the form b + c√3, where b and c are integers to be found.
(3)
(Total 4 marks)

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14. (a) Expand and simplify (4 + √3)(4 – √3).

(2)

26
(b) Express in the form a + b√3, where a and b are integers.
4+√3
(2)
(Total 4 marks)

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15. (a) Write √45 in the form a√5, where is an integer.

(1)

2(3 + √ 5)
(b) Express in the form b + c√5, where b and c are integers.
(3 − √ 5)
(5)
(Total 6 marks)

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Algebra: Surds and Indices – Questions 5

 PhysicsAndMathsTutor.com
1
16. (a) Write down the value of 8 3 .
(1)
−2
(b) Find the value of 8 3 .
(2)
(Total 3 marks)

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1
17. (a) Write down the value of 16 2 .
(1)
−3
(b) Find the value of 16 2 .
(2)
(Total 3 marks)

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18. Given that

f(x) = x2 – 6x + 18, x ≥ 0,

(a) express f(x) in the form (x – a)2 + b, where a and b are integers.
(3)

The curve C with equation y = f(x), x ≥ 0, meets the y-axis at P and has a minimum
point at Q.

(b) Sketch the graph of C, showing the coordinates of P and Q.

(4)

The line y = 41 meets C at the point R.

(c) Find the x-coordinate of R, giving your answer in the form p + q√2, where p and q
are integers.
(5)
(Total 12 marks)

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Algebra: Surds and Indices – Questions 6

 PhysicsAndMathsTutor.com

19. Solve the equation 21 – x = 4x.

(Total 3 marks)

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20. Giving your answers in the form a + b√2, where a and b are rational numbers, find

(a) (3 – √8)2,
(3)

1
(b) .
4− 8
(3)
(Total 6 marks)

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21. The curve C with equation y = f(x) is such that

dy 12
= 3√x + , x > 0.
dx √x

dy
(a) Show that, when x = 8, the exact value of is 9√2.
dx
(3)

The curve C passes through the point (4, 30).

(b) Using integration, find f(x).

(6)
(Total 9 marks)

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Algebra: Surds and Indices – Questions 7

 PhysicsAndMathsTutor.com

22. (a) Given that 3x = 9y −1, show that x = 2y – 2.

(2)

(b) Solve the simultaneous equations

x = 2y – 2,

x2 = y2 + 7.
(6)
(Total 8 marks)

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23. Express
2 2 2 3
− ,
3 −1 2 +1

in the form p 6 + q 3 + r 2 , where the integers p, q and r are to be found.

(Total 4 marks)

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24. Find the value of

1
(a) 81 2 ,
(1)
3
(b) 81 4 ,
(2)
−3
(c) 81 4 .
(1)
(Total 4 marks)

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Algebra: Surds and Indices – Questions 8

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

1. ( )
75 − 27 =5 3 − 3 3

= 2 3 A1
Note
for 5√3 from √75 or 3√3 from √27 seen anywhere
or k = 2, x = 3
A1 for 2 3 ; allow 12 or
allow k = 1, x = 12
Some Common errors
75 – 27 = 48 leading to 4 3 is M0A0
25 3 – 9 3 = 16 3 is M0A0
[2]

Algebra: Surds and Indices – Mark Schemes 9

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

2. (a) (7 + √ 5)(3 – √ 5)= 21– 5 + 3√ 5 – 7√ 5 Expand to get 3 or 4 terms

= 16, – 4√ 5 (1st A for 16, 2nd A for – 4√5) A1 A1 3

(i.s.w. if necessary, e.g.
16– 4√ 5 → 4 – √ 5)
Note
Allowed for an attempt giving 3 or 4 terms, with at least 2
correct (even if unsimplified).

e.g. 21– √ 52 + √15 scores

Answer only: 16 – 4√ 5 scores full marks
One term correct scores the M mark by implication,
e.g. 26 – 4√ 5 scores A0 A1

7+ 5 3– 5
(b) × (This is sufficient for the M mark)
3+ 5 3– 5
Correct denominator without surds, i.e. 9 – 5 or 4 A1
4 – √ 5 or 4 –1√ 5 A1 3
Note
Answer only: 4 – √ 5 scores full marks
One term correct scores the M mark by implication,
e.g. 4 + √ 5 scores A0 A0
16 – √ 5 scores A0 A0
Ignore subsequent working, e.g. 4 – √ 5 so a = 4, b = 1
Note that, as always, A marks are dependent upon the
7+ 5 3+ 5 .........
preceding M mark, so that, for example, × = is
3+ 5 3– 5 4
M0 A0.
Alternative
(a + b√ 5)(3 + √ 5)= 7 + √ 5, then form simultaneous equations
in a and b.
Correct equations: 3a + 5b = 7 and 3b + a = 1 A1
a=4 and b = –1 A1
[6]

Algebra: Surds and Indices – Mark Schemes 10

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

3. (a) (3 7 )2
= 63 B1 1
Note
B1 for 63 only

(b) (8 + 5 )(2 – 5 )=16 – 5 + 2 5 –8 5

= 11, – 6 5 A1, A1 3
Note
for an attempt to expand their brackets with ≥ 3 terms correct.
They may collect the 5 terms to get 16 – 5 – 6 5

Allow – 5 × 5 or – ( 5 ) or –
2
25 instead of the - 5

These 4 values may appear in a list or table but they should have
minus signs included
The next two marks should be awarded for the
final answer but check that correct values follow
from correct working. Do not use ISW rule

1st A1 for 11 from 16 – 5 or – 6 5 from – 8 5 + 2 5

2nd A1 for both 11 and – 6 5.

S.C – Double sign error in expansion
For 16 – 5 – 2 5 + 8 5 leading to 11 + … allow one mark
[4]

Algebra: Surds and Indices – Mark Schemes 11

 Algebra – Surds and Indices PhysicsAndMathsTutor.com
1
32 = 25 or 2048 = 211,
1
4. 2 =2 2 or 2048 = (2048) 2 B1, B1
11  1 
a=  or 5 or 5.5  B1
2  2 
Note

1st B1 for 32 = 25 or 2048 = 211

This should be explicitly seen: 32 2 = 2 a followed by 2 5 2 = 2 a
is OK Even writing 32 × 2 = 25 × 2(=26) is OK but simply writing
32 × 2 = 26 is NOT
1 1
2nd B1 for 2 2 or (2048) 2 seen. This mark may be implied
11
3rd B1 for answer as written. Need a = ... so 2 2 is B0
11  1 
a=  or 5 or 5.5  with no working scores full marks.
2 2 
If a = 5.5 seen then award 3/3 unless it is clear that the value
follows from totally incorrect work.
Part solutions: e.g. 2 5 2 scores the first B1.
Special case:
1 1
If 2 =2 2 is not explicitly seen, but the final answer includes ,
2
1 1
e.g. a = 2 , a = 4 , the second B1 is given by implication.
2 2
[3]

Algebra: Surds and Indices – Mark Schemes 12

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

5. (a) [(3 – 4 ]
x ) 2 = 9 – 12 x – 12 x + (– 4) 2 x
1 1
–
9x 2 + 16 x 2 – 24 A1, A1 3
Note

for an attempt to expand (3 – 4 x )2 with at least 3 terms

correct– as printed or better
Or 9 – k x + 16 x (k ≠ 0). See also the MR rule below
1
1st A1 for their coefficient of
(±)
x = 16. Condone writing (±) 9 x 2

1
–
instead of 9x 2

2nd A1 for B = – 24 or their constant term = – 24

3 1
9 – 16 –
(b) f’(x) = – x 2 , + x 2 A1, A1ft 3
2 2
Note
for an attempt to differentiate an x term x n → x n –1
3
9
1st A1 for – x
–
2 and their constant B differentiated to zero. NB
2
3
1 –
– × 9x 2 is A0
2
1
2nd A1ft follow through their Ax 2 but can be scored without a value for
1
A –2
A, i.e. for x
2

9 1 16 1 1 16 5
(c) f ′(9) = – × + × =– + = A1 2
2 27 2 3 6 6 2
Note
for some correct substitution of x = 9 in their expression for f′(x)
k
±
including an attempt at (9) 2 (k odd) somewhere that leads to
1
some appropriate multiples of or 3
3
15 45 135 67.5
A1 accept or any exact equivalent of 2.5 e.g. , or even
6 18 54 27
2
(3 – k x )
Misread (MR) Only allow MR of the form N.B. Leads to
x
k 2 –1
answer in (c) of Score as M1A0A0, M1A1A1ft, M1A1ft
6
[8]

Algebra: Surds and Indices – Mark Schemes 13

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

6. (a) 5 (± 5 is B0) B1 1

2
1  1 
(b) 2
or  
( their 5)  their 5 
1 1
= or 0.04 (± is A0) A1 2
25 25

Note
follow through their value of 5. Must have reciprocal and square.
1
5–2 is not sufficient to score this mark, unless follows this.
52
A negative introduced at any stage can score the but not the A1,
– 23  1 1
2
e.g. 125 = –  = scores A0
 5 25
– 23  1
2
1
125 = –  =− scores A0.
 5 25
Correct answer with no working scores both marks.
1 1
Alternative: = 2
or 1
(reciprocal and the correct
3
125 (125 2 ) 3

number squared)
 1 
 = 
3
 15625 
1
= A1
25
[3]

Algebra: Surds and Indices – Mark Schemes 14

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

49 – 22
2
7. 7 + 2 7 – 2 7 – 2 2 , or 7 – 4 or an exact equivalent such as
3 A1
Note
for an expanded expression. At worst, there can be one wrong term and
one wrong sign, or two wrong signs.
e .g. 7 + 2 7 – 2 7 – 2 is (one wrong term – 2)
7+ 2 7 + 2 7 4 is (two wrong signs + 2 7 and + 4)
7+ 2 7 + 2 7 + 2 7 + 2 is (one wrong term + 2, one wrong sign + 2 7
7 + 2 7 – 2 7 + 4 is (one wrong term 7 , one wrong sign + 4)
7 + 2 7 – 2 7 – 2 is M0 (two wrong terms 7 and – 2)

7 + 14 – 14 – 4 is M0 (two wrong terms 14 and – 14

If only 2 terms are given, they must be correct, i.e. (7 – 4) or an equivalent
unsimplified version to score
The terms can be seen separately for the
Correct answer with no working scores both marks.
[2]

Algebra: Surds and Indices – Mark Schemes 15

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

3
(a) or p = (Not 2 x x ) B1
3
8. 2x 2

– x or – x1 or q = 1 B1 2
Note

1st B1 for p = 1.5 or exact equivalent

2nd B1 for q = 1

 dy  3 1
(b) 3
 =  20 x + 2 × x 2 – 1
 dx  2
1
20 x 3
+ 3x 2 –1 A1A1ftA1ft 4

Note

for an attempt to differentiate xn→ xn–1 (for any of the 4 terms)

1st A1 for 20x3 (the – 3 must ‘disappear’)
1
2nd A1ft or 3 x . Follow through their p but they must be
for 3x 2
differentiating 2xp, where p is a fraction, and the coefficient
must be simplified if necessary.

3rd A1ft for –1 (not the unsimplified – x0), or follow through for correct
differentiation of their – xq (i.e. coefficient of xq is – 1).
If ft is applied, the coefficient must be simplified if necessary.
a
‘Simplified’ coefficient means where a and b are integers with no
b
common
factors. Only a single + or – sign is allowed (e.g. – – must be
replaced by +).
If there is a ‘restart’ in part (b) it can be marked independently
of part (a), but marks for part (a) cannot be scored for work seen in (b).

Multiplying by x : (assuming this is a restart)

e.g. y = 5x4 x – 3 x + 2x2 – x 2

3

 dy  45 7 2 3 – 12 3 1
 =  x – x + 4 x – x 2 scores A0 A0 (p not a
 d x  2 2 2
fraction) A1ft.
Extra term included: This invalidates the final mark.

e.g. y = 5x4 – 3 + 2x2 – x – x

3 1
2 2

Algebra: Surds and Indices – Mark Schemes 16

 Algebra – Surds and Indices PhysicsAndMathsTutor.com
 dy  3 1 1 –1
 = 20 x + 4 x – x 2 – x 2 scores A1 A0 (p not a
3

 dx  2 2
fraction) A0.
Numerator and denominator differentiated separately:
For this, neither of the last two (ft) marks should be awarded.
Quotient/product rule:
Last two terms must be correct to score the last 2 marks. (If the M mark
has not already been earned, it can be given for the quotient/product
rule attempt.)
[6]

Algebra: Surds and Indices – Mark Schemes 17

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

9. (a) 2 B1 1
Negative answers:
Allow –2. Allow ±2. Allow ‘2 or –2’.

(b) x9 seen, or (answer to (a))3 seen, or (2x3)3 seen.

8x9 A1 2

M: Look for x9 first… if seen, this is

If not seen, look for (answer to (a))3, e.g. 23… this would score
even if it does not subsequently become 8. (Similarly for other
answers to (a)).

In (2x3)3, the 23 is implied, so this scores the M mark.

Negative answers:

Allow ±8x9. Allow ‘8x9 or –8x9’.

N.B. If part (a) is wrong, it is possible to ‘restart’ in part (b) and to score full marks in
part (b).
[3]

Algebra: Surds and Indices – Mark Schemes 18

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

(5 − 3 ) (2 − 3 )
10. ×
(2 + 3 ) (2 − 3 )

10 − 2 3 − 5 3 + ( 3 ) 2  10 − 7 3 + 3 
= = 
...  ... 
 
 13 − 7 3 
(= 13 − 7 3 )  Allow 13 (a = 13) A1
 1 
 

− 7 3 (b = −7) A1 4

1st M: Multiplying top and bottom by (2 – 3 ). (As shown above is sufficient).

2nd M: Attempt to multiply out numerator (5 – 3 ) (2 – 3 ). Must have at

least 3 terms correct.
Final answer: Although ‘denominator = 1’ may be implied, the 13 − 7 3
must obviously be the final answer (not an intermediate step), to score
full marks. (Also M0 A1 A1 is not an option).

The A marks cannot be scored unless the 1st M mark has been scored,
but this 1st M mark could be implied by correct expansions of both
numerator and denominator.
It is possible to score M0 A1 A0 or M0 A0 A1 (after 2 correct terms
in the numerator).

Special case: If numerator is multiplied by (2 + 3 ) instead of (2 − 3 ) ,

the 2nd M can still be scored for at least 3 of these terms correct:
10 − 2 3 + 5 3 − ( 3 ) 2 .
The maximum score in the special case is 1 mark: M0 A0 A0.
Answer only: Scores no marks.
Alternative method:
5 − 3 = (a + b 3 )(2 + 3 )
(a + b 3 )(2 + 3 ) = 2a + a 3 + 2b 3 + 3 At least 3 terms correct.
5 = 2a + 3b
–1 = a + 2b a =… or b =… Form and attempt to solve
simultaneous equations.
a = 13, b = –7 A1, A1
[4]

Algebra: Surds and Indices – Mark Schemes 19

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

11. 9 – 5 or 32 + 3 5 − 3 5 − 5 × 5 or 32 – 5 × 5 or 32 – ( 5 )2
=4 A1cso 2
for an attempt to multiply out. There must be at least 3 correct terms.
Allow one sign slip only, no arithmetic errors.

e.g. 32 + 3 5 − 3 5 + ( 5 ) 2 is M1A0
32 + 3 5 + 3 5 − ( 5 ) 2 is M1A0 as indeed is 9 ± 6 5 – 5

BUT 9 + 15 − 15 – 5(= 4) is M0A0 since there is more than a sign error.

6 + 3 5 − 3 5 – 5 is M0A0 since there is an arithmetic error.
If all you see is 9 ± 5 that is but please check it has not come from
incorrect working.
Expansion of (3 + 5 )(3 + 5 ) is M0A0
A1cso for 4 only. Please check that no incorrect working is seen.
Correct answer only scores both marks.
[2]

Algebra: Surds and Indices – Mark Schemes 20

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

12. (a) Attempt 3 8 or 3

(8 4 )
= 16 A1 2
4
for: 2 (on its own) or (2 3
)3 or 3 8 or (3 8 ) 4 or 24
or 3 8 4 or 3
4096
1
83 or 512 or (4096) 3 is M0
A1 for 16 only

1 1
(b) 5x 3 5, x 3 B1, B1 2

1st B1 for 5 on its own or × something.

4
1
5x 3
So e.g. is B1 But 53 is B0
x
An expression showing cancelling is not sufficient
(see first expression of QC0184500123945 the mark
is scored for the second expression)
1
2nd B1 for x3
Can use ISW (incorrect subsequent working)
4
e.g.5x 3scores B1B0 but it may lead to 3
5x 4 which
we ignore as ISW.
Correct answer only scores full marks in both parts.
[4]

13. (a) 6√3 (a = 6) B1 1

±6√3 also scores B1.

(b) Expanding (2 – √3)2 to get 3 or 4 separate terms

7, –4√3 (b = 7, c = –4) A1, A1 3
The 3 or 4 terms may be wrong.

1st A1 for 7, 2nd A1 for –4√3.

Correct answer 7 – 4 √3 with no working scores all 3 marks.
7 + 4 √3 with or without working scores A1 A0.
Other wrong answers with no working score no marks.
[4]

Algebra: Surds and Indices – Mark Schemes 21

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

14. (a) 16 + 4 3 − 4 3 − ( 3 ) 2 or 16 − 3

= 13 A1c.a.o 2
For 4 terms, at least 3 correct
e.g. 8 + 4 3 − 4 3 − ( 3 ) 2 or 16 ± 8 3 − ( 3 ) 2 or
16 + 3
42 instead of 16 is OK
(4 + 3 )(4 + 3 ) scores M0A0

26 4− 3
(b) ×
4+ 3 4− 3

26(4 − 3 )
= = 8 − 2 3 or 8 + (−2) 3 or a = 8 and b = −2 A1 2
13
For a correct attempt to rationalise the denominator
Can be implied
− 4+ 3
NB is OK
− 4+ 3
[4]

15. (a) 3√5 (or a = 3) B1 1

2(3 + √ 5) (3 + √ 5)
(b) ×
(3 − √ 5) (3 + √ 5)

(3 – √5)(3 + √5) = 9 – 5 (= 4) (Used as or intended as denominator) B1

(3 + √5)(p ± q √5) = ... 4 terms (p ≠ 0, q ≠ 0) (Independent)
or (6 + √5)(p ± q√5) = ... 4 terms (p ≠ 0, q ≠ 0)
[Correct version: (3 + √5)(3 + √5) = 9 + 3√5 + 3√5 + 5,or double this.]
2(14 + 6 √ 5)
= 7 + 3√5 1st A1: b = 7, 2nd A1: c = 3 A1 A1 5
4
[6]

2nd M mark for attempting (3 + √5)(p + q √5) is generous. Condone errors.

Algebra: Surds and Indices – Mark Schemes 22

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

16. (a) 2 B1 1
Penalise ±

2
− 1 1 1 1
(b) 8 3 = or or or
3
64 (a) 2 3
8 2 2
83
Allow ±
1
= or 0.25 A1 2
4
for understanding that “-“ power means reciprocal
2
1
8 3 = 4 is M0A0 and – is M1A0
4
[3]

17. (a) 4 (or ± 4) B1

3 3
− 1
(b) 16 2 = 3
and any attempt to find 16 2
16 2
1 1
(or exact equivalent, e.g. 0.015625) (or ± ) A1 3
64 64
[3]

18. (a) x2 – 6x + 18 = (x – 3)2, +9 B1, A1 3

30 y

(b) 20

10

x
–1 1 2 3 4 5 6

“U”-shaped parabola
Vertex in correct quadrant A1ft
P: (0, 18) (or 18 on y-axis) B1
Q: (3, 9) B1ft 4

(c) x2 – 6x + 18 = 41 or (x – 3)2 + 9 = 41
Attempt to solve 3 term quadratic x = …
6 ± 36 − (4 × −23)
x= (or equiv.) A1
2
√128 = √64 × √2 (or surd manipulation 2a = 2 a )
3 + 4√2 A1 5
[12]

Algebra: Surds and Indices – Mark Schemes 23

 Algebra – Surds and Indices PhysicsAndMathsTutor.com

19. 4 = 22 (or log 4 = 2log 2) B1

Linear equation in x: 1 – x = 2x
1
x= A1 3
3
[3]

20. (a) √8 = √4√2 = 2√2 (seen or implied) B1

(3 – √8)(3 – √8) = 9 – 6√8 + 8 = 17 – 12√2 A1 3

1 4 + √8 4 + √8 1 1
(b) × , = = + √2 A1 3
4 − √8 4 + √8 16 − 8 2 4
1
Allow (2 + √ 2) or equiv. (in terms of √2)
4
[6]

21. (a) √8 = 2√2 seen or used somewhere (possibly implied). B1

12 12 8 12 12 2
= or =
8 8 2 2 4
6
Direct statement, e.g. = 3 2 (no indication of method) is M0.
2
dy 12
At x = 8, =3√8 + = 6√2 + 3√2 = 9√2 (*) A1 3
dx 8

3 1
3x 12 x 2
2
(b) Integrating: + (+C ) (C not required) A1 A1
3 ( ) ( )
2
1
2
3 1
3× 4 2 12 × 4 2
At (4, 30), + + C = 30 (C required)
3
2
( ) 1
2
( )
3 1
(f(x) =) 2 x 2 + 24 x 2 , –34 A1, A1 6
[9]

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22. (a) 3x = 32(y–1) x = 2 (y −1) (*) A1

(b) (2y − 2)2 = y2 + 7, 3y2 − 8y − 3 = 0 A1

(3y + 1)(y − 3) = 0, y = … (or correct substitution in formula)
1
y=− , y=3 A1
3
8
x=− , x=4 A1 ft
3
[8]

23. Rationalising either surd

2 2 ( 3 + 1) 2 3 ( 2 − 1)
− A1, A1
2 1
−√6 + 2√3 + √2 (or p = −1, q = 2, r = 1) A1 4
[4]

24. (a) 9 B1 1

33 = 27
1
(b) 81 4 = 3 A1 2
1
(c) B1 ft 1
27
[4]

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1. This was a successful starter to the paper with very few candidates failing to attempt it
and many securing both marks. The √27 was usually written as 3√3 but 5√5 and 3√5
were common errors for √75. The most common error though was to subtract 27 from
75 and then try and simplify √48 which showed a disappointing lack of understanding.

2. This question was answered very well, with many candidates scoring full marks.
Mistakes in part (a) were usually from incorrect squaring of the 5 term, sign errors or
errors in collecting the terms. In part (b), the method for rationalising the denominator
was well known and most candidates, whether using their answer to part (a) or not,
proceeded to a solution. A common mistake, however, was to divide only one of the
terms in the numerator by 4.

3. Some candidates could not square the surd terms correctly but nearly everyone
attempted this question and most scored something.
In part (a) some failed to square the 3 and an answer of 21 was fairly common, others
realised that the expression equalled 9 × 7but then gave the answer as 56. A few
( )2
misread the question and proceeded to expand 3 + 7 . In part (b) most scored a mark
for attempting to expand the brackets but some struggled here occasionally adding 8 + 2
instead of multiplying. Those with a correct expansion sometimes lost marks for
careless errors, − 8 5 + 2 5 + = 6 5, and a small number showed how fragile their
understanding of these mathematical quantities was by falsely simplifying a correct
answer of 11 − 6 5 to 5 5.

4. Many candidates could not deal with this test of indices. Two simple properties of
indices were required: that a square root leads to a power of 12 and the rule for adding
the powers when multiplying. Those who identified these usually made good progress
but the remainder struggled. Some re-wrote 32 2 as 64 and then obtained a = 3 whilst
others did obtain 2048 but usually failed to identify 2048 as 211. A number of
candidates tried to use logs but this approach was rarely successful.

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5. Most candidates made a good attempt at expanding the brackets but some struggled
1
with − 4 x × −4 x with answers such as − 4 x or ± 16 x or ± 16 x 4 being quite common.
x
The next challenge was the division by x and some thought that = 1. Many, but not
x
all, who had difficulties in establishing the first part made use of the given expression
and there were plenty of good attempts at differentiating. Inevitably some did not
interpret f '(x) correctly and a few attempted to integrate but with a follow through mark
here many scored all 3 marks. In part (c) the candidates were expected to evaluate
3 1
− −
9 correctly and then combine the fractions - two significant challenges but
2 or 9 2

many completed both tasks very efficiently.

6. Many candidates answered both parts of this question correctly. In part (b), however,
some did not understand the significance of the negative power. Others, rather than
using the answer to part (a), gave themselves the difficult, time-wasting task of squaring
the 125 and then attempting to find a cube root. Negative answers (or ±) appeared
occasionally in each part of the question.

7. Most candidates completed this question successfully, either by expanding the brackets
to find four terms or by recognising the difference of squares and writing down 7 – 4 =
3 directly. Common wrong answers included 11 + 4 7 , from ( 7 + 2)( 7 + 2), and 5,
from 7 + 2 7 – 2 7 – 2. Mistakes such as 7 × 2 = 14 were rarely seen.

8. Good candidates generally had no difficulty with the division in part (a) of this question,
but others were often unable to cope with the required algebra and produced some very
confused solutions. A common mistake was to “multiply instead of divide”, giving
5
2x ÷ x =2
2x 2 , and sometimes √x was interpreted as x–1. Examiners saw a wide
variety of wrong answers for p and q.
Most candidates were able to pick up at least two marks in part (b), where follow-
through credit was available in many cases. While the vast majority used the answers
from part (a), a few differentiated the numerator and denominator of the fraction term
separately, then divided.

9. Although many candidates obtained the correct answer in part (a), they were usually
3
unable to use this to find (16 x 12 ) 4 in part (b). Even very good candidates tended to
score only one mark here, with the most common incorrect answers being 16x9 and
51
12x9. Other wrong answers included those in which powers had been added to give x 4 .
Very confused algebra was sometimes seen.

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10. There were many completely correct solutions to this question. The majority of
candidates knew the correct method of multiplying the numerator and denominator by
(2 – √3) and many were correct in the arithmetic manipulation. Some multiplied
incorrectly by (2 + √3) or by (5 + √3). A number of candidates were unable to square √3
correctly and it was disappointing to see marks lost though careless arithmetic. Only a
small minority of candidates had no idea of how to start.

11. This proved an easy starter for most candidates. Some identified this as a difference of
two squares, and simply wrote down 9 – 5, but most opted to multiply out and sign slips
spoilt some answers with 9+5 appearing quite often. Others thought that 3 5 = 15 and
some wrote 32 = 6.

12. There were many correct responses to both parts of this question. In part (a) some
reached 3 4096 but could not simplify this expression but most managed 3 8 = 2 and
( ) 3
usually went on to give the final answer of 16. A few attempted 4 8 but most
interpreted the notation correctly. Part (b) revealed a variety of responses from those
whose grasp of the basic rules of algebra is poor. Most simplified the numerical term to
1 4
5 but often they seemed to think the x terms “disappeared” and answers of 5 3 or 5 3
4
were common. Dealing with the x term proved quite a challenge for some and (5 x) 3
was a common error. Some candidates tried to “simplify” a correct answer, replacing
1
5 x 3 with 3 5 x . On this occasion the examiners ignored this subsequent working but such
a misunderstanding of the mathematical notation used in AS level mathematics is a
legitimate area to be tested in future.

13. It was encouraging to see most candidates factorizing the quadratic expression in order
to find the critical values for the inequality. Sometimes the critical values had incorrect
signs, despite the factorization being correct, but the most common error was still a
failure to select the outside region. Some candidates struggled with the correct symbolic
notation for the answer and –2 > x > 9 was occasionally seen.
A few candidates chose to use the formula or completing the square to find the critical
values, these approaches are less efficient in this case and often gave rise to arithmetic
errors.

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14. This question was generally answered well although there was the usual crop of
arithmetic and sign errors especially in part (a) where some candidates struggled to
simplify ( 3 ) . In part (b) most knew how to start the problem, although a few
2

4+ 3
multiplied by . It was disappointing to see how many candidates multiplied out
4+ 3
the numerator first and then divided by 13, often forgetting to divide one of the two
terms by 13.

15. The surd simplification in part (a) was completed successfully by most candidates,
although 9√5 was sometimes seen instead of 3√5. In part (b), candidates often showed
competence in the process of rationalising the denominator, but sometimes made
mistakes in multiplying out their brackets, with frequent mishandling of terms involving
28 + 12 √ 5
√5. It was disappointing to see a few candidates proceeding correctly to and
4
then dividing only one of the terms, to give, for example, 7 + 12√5. Not surprisingly,
those who were unfamiliar with rationalising the denominator usually scored no marks
in part (b).

16. This was a successful starter to the paper and nearly all the candidates were able to
make some progress. Part (a) was rarely incorrect although sometimes the answer was
given as ±2. The negative power led to a number of errors in part (b). Some thought that
−2 3
8 3 = 8 2 whilst others thought that a negative answer (usually – 4) was appropriate.
Most knew that the negative power meant a reciprocal and the correct answer was often
seen.

17. There were many completely correct answers to this question. The vast majority of
candidates knew that a square root was required in part (a), but part (b) caused a few
more difficulties, with the negative power not always being interpreted to mean a
reciprocal. Some cubed 16 before finding a square root, making unnecessary work for
3
themselves, while others evaluated 16 2 as 12 or 48.

18. Most candidates were familiar with the method of “completing the square” and were
able to produce a correct solution to part (a). Sketches in part (b), however, were often
disappointing. Although most candidates knew that a parabola was required, the
minimum point was often in the wrong position, sometimes in the fourth quadrant and
sometimes at (3, 0). Some candidates omitted part (c), but most knew what was required
and some very good, concise solutions were seen. Those who used the answer to part (a)
and formed the equation ( x − 3) 2 + 9 = 41 were able to proceed more easily to an answer
from x = 3 ± 32 . Some candidates found difficulty in manipulating surds and could not
cope with the final step of expressing √32 as 4√2.

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19. Most candidates completed this question successfully, although marks were sometimes
lost through careless algebra. The most popular method was to write 4 x as (22 ) x , then
to equate powers and solve the resulting linear equation. Other approaches, including
the use of logarithms, were occasionally seen.

20. Although good candidates often produced fully correct solutions to this question, others
lacked confidence in using surds and found the manipulation difficult. In both parts of
the question, a significant number of candidates left their answer in terms of √8, failing
to demonstrate the simplification to 2√2. In part (a), most earned a method mark for
their attempt to square the bracket, but numerical and sign errors were common. In part
(b), however, many candidates did not realise that they needed to (or did not know how
to) rationalise the denominator, and therefore made little progress. It was disappointing
1 1 1
to see attempts starting with = − .
4 − √8 4 √8

21. The manipulation of surds in part (a) was often disappointing in this question. While
most candidates appreciated the significance of the “exact value” demand and were not
tempted to use decimals from their calculators, the inability to rationalise the
denominator was a common problem. Various alternative methods were seen, but for all
of these, since the answer was given, it was necessary to show the relevant steps in the
working to obtain full marks.
Integration techniques in part (b) were usually correct, despite some problems with
fractional indices, but the lack of an integration constant limited candidates to 3 marks
out of 6. Sometimes the (4, 30) coordinates were used as limits for an attempted
“definite integration”.

22. In part (a), most candidates were able to use 9 = 3 to show convincingly that x = 2y - 2.
Just a few used logarithms, usually correctly, to complete the proof. Part (b), however,
proved to be a good test of algebraic competence. Forming an appropriate equation in y
was the important step, and candidates who thought that x = 2y - 2 implied x = 4y – 4
(or similar) oversimplified the y equation to a two-term quadratic and limited
themselves to a maximum of 2 marks.
Otherwise, apart from slips in expanding (2y – 2) , many went on to achieve a fully
correct solution. It was unfortunate, however, that some candidates solved the quadratic
in y but gave their answers as values of x, substituting back to find “y”. Others, having
solved for y, omitted to find the corresponding values of x.

23. No Report available for this question.

24. No Report available for this question.

Algebra: Surds and Indices – Examiner Reports 30