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Quadratic Equations
All types, factorising, equation, completing the square
165 minutes
151 marks
Page 1 of 53
Q1. (a) Factorise x2 + 5x − 24
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Answer ...................................................................... (2)
(b) Solve x2 + 5x − 24 = 0
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Answer ...................................................................... (1)
(Total 3 marks)
Q2. Use the quadratic formula to solve
6x2 + 5x − 3 = 0
Give your answers to 2 decimal places.
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Answer ............................... and ............................... (Total 3 marks)
Page 2 of 53
Q3. Solve the quadratic equation 3x2 + x – 5 = 0
Give your answers to 3 significant figures.
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Answer ...................................................................... (Total 3 marks)
Q4. The graph of y = x2 + 2x − 3 is drawn on the opposite page.
Page 3 of 53
Draw an appropriate straight line on the graph to work out the approximate solutions of
x2 + x − 3 = 0
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Answer ......................................................................
y = x2 + 2x − 3
(Total 3 marks)
Page 4 of 53
Q5. Solve the equation 2x2 + 8x + 5 = 0
Give your answers to 2 decimal places.
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Answer ...................................................................... (Total 3 marks)
Page 5 of 53
Q6. Work out the value of x.
Not drawn accurately
You must show your working.
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x = ............................................................................. (Total 5 marks)
Page 6 of 53
Q7. A car and a lorry complete the same 240 mile journey without stopping.
The average speed of the car is x mph. The average speed of the lorry is 12 mph slower than the car.
The lorry takes 1 hour longer than the car.
Use an algebraic method to work out the average speed of the car.
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Answer .............................................................. mph (Total 6 marks)
Page 7 of 53
Q8. Solve x2 + 8x + 6 = 0 by completing the square.
Give your answer in the form a ± , where a and b are integers.
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Answer ...................................................................... (Total 4 marks)
Q9. Solve the equation 3x2 + 4x − 10 = 0
Give your answers to 2 significant figures.
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Answer ...................................................................... (Total 3 marks)
Page 8 of 53
Q10. Solve the quadratic equation
6x2 + 2x − 5 = 0
Give your answers to 2 decimal places.
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Answer ...................................................................... (Total 3 marks)
Q11. Solve the equation x2 − 5 = 0
Give your answers to 1 decimal place.
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Answer ............................... and .............................. (Total 2 marks)
Q12. (a) Show that x2 − 8x + 20
can be written in the form (x − a)2 + a
where a is an integer.
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......................................................................................................................... (3)
Page 9 of 53
(b) Hence explain how you know that x2 − 8x + 20 is always positive.
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......................................................................................................................... (2)
(Total 5 marks)
Q13. The diagram shows a rectangle.
The area of the rectangle is 90 cm2.
Set up and solve a quadratic equation to work out the value of x.
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x = ....................................................................... cm (Total 5 marks)
Q14. Solve the equation 5x2 + 14x − 24 = 0
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Answer ...................................................................... (Total 3 marks)
Page 10 of 53
Q15. Two numbers, a and b, are combined using the operation in the following way.
a b= 2a2 − 7a − b + b2
Work out all solutions of the equation x 3 = 0
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Answer ...................................................................... (Total 4 marks)
Q16.
simplifies to 7x2 + 15x − 18 = 0
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Show that
Page 11 of 53
Q17. Given that x2 + ax + b ≡ (x − 7)2 − a
work out the values of a and b.
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Answer a = ............................ b = ............................. (Total 3 marks)
Q18. Katy is using the quadratic formula to solve a quadratic equation.
After correctly substituting the values, she writes
(a) What is the quadratic equation Katy is trying to solve?
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Answer ............................................... (3)
(b) Explain why Katy will not be able to find any solutions to the equation.
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......................................................................................................................... (1)
(Total 4 marks)
Page 12 of 53
Q19. This right-angled triangle has sides of lengths (x – 2) cm, (x + 5) cm and 10 cm.
Calculate the value of x. Give your answer to an appropriate degree of accuracy.
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Answer ........................................... cm (Total 5 marks)
Q20. (a) Expand and simplify 2x (x + 6) + 3x(x – 5)
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Answer .................................................. (3)
Not drawn accurately
2
Page 13 of 53
(b) Factorise fully 3mh – 15m h
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Answer .................................................. (2)
(c) Simplify fully 4rs × 5r s
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Answer .................................................. (2)
(d) Solve 9x + 29x – 28 = 0
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Answer .................................................. (3)
(Total 10 marks)
2 2
2 3 4
2
Page 14 of 53
Q21. Solve the equation (2x – 3) = (x – 1)(x + 1)
Give your solutions to 2 decimal places.
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Answer ................................................ (Total 5 marks)
Q22. Solve the equation 3x – 5x – 7 = 0
Give your answers to 2 decimal places.
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Answer ....................................................................... (Total 3 marks)
2
2
Page 15 of 53
Q23. You are given the identity x – 12x + 10 ≡ (x + a) + b
Work out the values of a and b.
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Answer a = .......................... b = .......................... (Total 2 marks)
Q24. (a) (i) Factorise x – x – 2
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Answer .............................................. (2)
(ii) Hence, solve x – x – 2 = 0
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Answer .............................................. (1)
(b) Simplify
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Answer .............................................. (3)
(Total 6 marks)
2 2
2
2
Page 16 of 53
Q25. A trapezium has parallel sides of length (x + 1) cm and (x + 2) cm. The perpendicular distance between the parallel sides is x cm. The area of the trapezium is 10 cm2.
Find the value of x.
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Answer x = .............................................. cm (Total 5 marks)
Q26. The grid below shows graphs of a curve
y = x2 + 2x – 3
and 3 straight lines
y = x + 1
y = – x – 2
and y = – x + 2
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Page 17 of 53
You must use the graphs to answer the following questions.
(a) Write down a pair of simultaneous linear equations that have a solution
x = – , y =
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Answer ................................................. (1)
Page 18 of 53
(b) Write down and simplify a quadratic equation whose solutions are approximately – 3.3 or 0.3. You must show clearly how you obtain your answer.
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Answer ................................................. (2)
(c) Write down the approximate solutions to the equation x2 + x – 4 = 0.
You must show clearly how you obtain your answer.
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Answer ................................................. (2)
(Total 5 marks)
Q27. The diagram shows the graph of the equation y = x2 + px + q
The graph crosses the x-axis at A and B (2,0).
C (–3, –5) also lies on the graph.
Page 19 of 53
(a) Find the values of p and q.
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Answer p = ...................... q = ...................... (4)
(b) Hence work out the coordinates of A.
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Answer ( ........................ , ........................ ) (2)
(Total 6 marks)
Page 20 of 53
Q28. The grid below shows the graph of y = x2 + 3x – 2
(a) By drawing an appropriate straight line on the graph solve the equation
x2 + 3x – 3 = 0
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Answer ................................................. (2)
Page 21 of 53
(b) By drawing an appropriate straight line on the graph solve the equation
x2 + 2x – 1 = 0
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Answer ................................................. (3)
(Total 5 marks)
Q29. The diagram shows the graph of an equation of the form y = x2+ bx + c
Find the values of b and c. You must show your method.
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Answer b = ................. , c = ................... (Total 3 marks)
Page 22 of 53
Q30. Find the values of a and b such that
x2 – 10x + 18 = (x – a)2 + b
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Answer a = ..........................., b = ........................... (Total 2 marks)
Page 23 of 53
Q31. The diagrams show a rectangle and an L shape All the angles are right angles. All lengths are in centimetres. The shapes are equal in area.
Calculate the value of y.
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Answer ................................................. cm (Total 6 marks)
Page 24 of 53
Q32. (a) Find the values of a and b such that
x2 + 6x – 3 = (x + a)2 + b
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Answer a = ........................., b = ......................... (2)
(b) Hence, or otherwise, solve the equation
x2 + 6x – 3 = 0
giving your answers in surd form.
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Answer ................................................. (3)
(Total 5 marks)
Page 25 of 53
Q33. Solve the equation
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Q34. Solve
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Answer .................................................. (Total 6 marks)
Page 26 of 53
Q35. Solve − = 2
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Answer ..................................................................... (Total 7 marks)
Page 27 of 53
M1. (a) (x + a)(x + b) where ab = ± 24
M1
(x + 8)(x – 3) either order
A1
(b) (x =) – 8 and (x =) 3
ft their part (a) B1 ft
[3]
M2. (x =)
Allow one error M1
(x =)
A1
0.40 and −1.24 A1
[3]
M3.
Allow one error M1
oe
A1
1.14 and −1.47 SC2 for 1.14 or −1.47
SC1 for 1.135… or . −1.468… A1
[3]
M4. (x2 + 2x − 3) − (x2 + x − 3)
Or attempt to ‘balance’ equations M1
Page 28 of 53
y = x A1
− 2.3 and 1.3 ft if M awarded and their line drawn
A1ft [3]
M5.
Allow one error oe
M1
or Fully correct oe
A1
−0.78 and −3.22 SC2 for −0.78 or −3.22
SC1 for −0.775… or −3.224…
−0.775… and − 3.224… implies M1A1 A1
[3]
M6. (2x + 1)2 or (2x − 1)2 oe
8, 15, 17 seen M1
4x2 + 2x + 2x + 1
or 4x2 + 4x + 1
A1
4x2 − 2x − 2x + 1
2x + 1 = 17 or 2x − 1 = 15
or 4x2 − 4x + 1
A1
Page 29 of 53
(2x + 1)2 = (2x − 1)2 + 82 oe
or their (2x + 1)2 = their (2x − 1)2 + 82
2x = 16
or
4x2 + 2x + 2x + 1
= 4x2 − 2x − 2x + 1 + 64
or 8x = 64 M1
8 Do not accept 8 without working
A1 [5]
M7. or
x − 12 = or x =
M1
and
x − 12 = and x =
M1dep
− = 1
− 12 =
M1dep
240 x − 240( x − 12) = x ( x − 12)
240(t + 1) − 12 t(t + 1) = 240t
or 2880 = x 2 − 12 x
or t2 + t − 20 = 0
or x2 − 12x − 2880 = 0 oe
M1 dep
Page 30 of 53
(x + 48)(x − 60)
(t + 5)(t − 4)
or correct use of formula or correct use of formula
or (x − 6 )2 −36 −2880 = 0
M1
60 SC2 for 60 from trial and error
A1 [6]
M8. (x + 4)
Must be in brackets M1
(x + 4)2 − 16 + 6
oe A1
x + 4 = ±
Allow their 10 to be negative M1dep
(x =) −4 ±
ft on one arithmetic error, but only if their 10 is positive
−4 + is 3 marks
Correct answer with no working is 3 marks maximum A1ft
Alternative method
(x) =
Must be correct use of formula M1
A1
(x =) −4 ± A1
[4]
Page 31 of 53
M9. Allow one error
M1
or (−4 ± 136 ) ÷ 6 Fully correct oe
M1 dep
(x =) 1.3 and −2.6 A1
[3]
M10. (x =)
Allow one error M1
(x =)
(x =) A1
0.76 and −1.09 M1
[3]
M11. x2 = 5
2.2(36 …) or or
M1
2.2 and −2.2 2.2 or 2.23… implies M1
A1 [2]
M12. use of (x − 4)2
M1
(x − 4)2 − 16 (+ 20)
A1
Page 32 of 53
(x − 4)2 − 16 + 20 = (x − 4)2 + 4
Strand (ii) Complete and correct algebraic explanation
Q1
Alternative method 1
use of (x − 4)2
M1
= x2 − 8x + 16
A1
(x − 4)2 + 4 = x2 − 8x + 20
Strand (ii) Complete and correct algebraic explanation
Q1
Alternative method 2
x2 − ax − ax + a2 (+a)
M1
a = 4 A1
Also 42 + 4 = 20
Strand (ii) Complete and correct algebraic explanation
Q1
(b) explains that a square is always positive (or zero) oe B1
and a positive number is added so is always positive oe B1
[5]
M13. (x + 4)(x − 5) (= 90) M1
x2 + 4x − 5x − 20 (= 90)
Allow 1 error M1
x2 − x − 110 (= 0)
Collecting their 4 terms and 90
dependent on 2nd M1 only
M1dep
Page 33 of 53
(x + 10)(x − 11)
(x + a)(x + b) where ab = ± their 110
Use of formula – allow one error M1
11 Note: 11 and − 10 implies M4A0
A1 [5]
M14. (5x ± a)(x ± b) (= 0) ab = 24 M1
(5x − 6)(x + 4) (= 0) A1
A1
Alternative method 1
Allow one substitution error but not a conceptual error M1
or better A1
oe A1
and −4
oe eg or 1.2
(x =) and −4
Page 34 of 53
Alternative method 2
(x + 1.4)2 − 1.96 − 4.8 (= 0)
Allow one numerical error M1
A1
(x =) 1.2 and −4
oe A1
[3]
M15. 2x2 − 7x − 3 + 32
M1
2x2 − 7x + 6
A1
(2x + a)(x + b) (= 0)
ab = ± their 6
Must be a quadratic in 2x2
Substitution in quadratic formula (if used) must be correct for M1 eg for 2x2 − 7x + 6 (= 0)
M1
1.5 and 2
oe
SC3 for 2x2 − 7x + 3 (= 0)
leading to answers of 0.5 and 3 A1
[4]
x + 1.4 = ±
Page 35 of 53
M16. 7x + = 9
M1 for equating two correct fractions
M1
7x(x + 2) + 10x = 9(x + 2)
oe M1 dep
7x2 + 14x + 10x = 9x + 18
or 7x2 + 14x + 10x − 9x − 18 = 0
A1 [3]
M17. x2 − 7x − 7x + 49 (−a) or x2 − 14x + 49 (−a)
M1
a = −14
a = −14 from no working or an error in the number term of the expansion implies M1 A1
A1
b = 63
ft for b = 35 from a = 14, if M mark earned A1 ft
Alternative method
Substitutes a value for x in the identity eg x = −1 gives b = 63 M1
a = −14 A1
b = 63 A1
[3]
or 7(x + 2) + 10 = (x + 2)
or 7 + − = 0
Also M1 for
Page 36 of 53
M18. (a) 2x2 – 7x + 9 = 0
oe –1 eeoo B2 fully correct expression
B3
(b) Reference to square root of negative number B1
[4]
M19. (x + 5)2 + (x – 2)2 = 102
x2 + 25 + x2 ± 4 = 100 implies M1
M1
x2 + 10x + 25 + x2 – 4x + 4 = 100
SC1 x2 + 10 x + 25 + x2 – 4 x + 4 = 10
A1
2x2 + 6x – 71 = 0
For rearranging into a 3 term quadratic = 0 or going to cts straight away.
M1 dep
Use of quadratic formula or completing the square to solve Any evidence of formula or cts gets M1 Allow use of graphical calculator
M1 dep
4.6, 4.64, 4.65 4.64(4..) for T & I
A1 [5]
M20. (a) 2x3 + 12x2 + 3x2 – 15x
Allow 1 error M1
2x3 + 12x2 + 3x2 – 15x
Fully correct A1
2x3 + 15x2 – 15x
Ignore factorising after final answer ft From 4 terms where simplification is possible A0 For fw eg, incorrect attempt to collect terms
A1 ft
Page 37 of 53
(b) 3mh(h – 5m) B1 For partial factorisation with two factors 3m(h2 – 5mh)
3h(mh – 5m2)
mh(3h – 15m) B2
(c) 20r 4s 6
B1 For 2 terms correct B2
(d) Attempt to factorise or attempt to use formula
(allow one error) (ax + b)(cx + d) where ac = 9 and bd = ± 28
M1
(9x – 7)(x + 4) or formula fully correct
A1
7 / 9 and –4 Accept 0.77 ... or 0.78
A1 [10]
M21. 4x2 – 12x + 9
or 4x2 – 6x – 6x + 9
3 correct terms out of 4 seen B1
x2 – 1
or x2 – x + x – 1
3 correct terms out of 4 seen B1
(their) 3x2 – 12x + 10 (= 0)
ft (their) 4x2 – 12x + 9 and (their) x2 – 1
oe M1
Page 38 of 53
ft (their) 3x2 – 12x + 10
Must be in the form ax2 + bx + c (= 0)
M1
2.82 and 1.18 A1
[5]
M22. Allow one error or [5 ± √59] ÷ 6
M1
(5 ± √ 109) ÷ 6 A1
2.57 and –0.91 SC1 2.3 and –0.45
A1 [3]
M23. (a =) –6 B1
(b =) –26 B1
[2]
M24. (a) (i) (x ± a)(x ± b)
ab = ± 2 M1
(x + 1)(x – 2) A1
(ii) (x = ) –1 and 2 ft Their brackets if M1 awarded
B1ft
Page 39 of 53
(b)
ft Their factorisation if M1 awarded in (a) M1
Numerator 3x + 2 + x – 2 ft Their factorisation
A1ft
or
No ft A1
[6]
M25. (Area =) x (x + 1 + x + 2)
oe (x + 1) + × x × (1) M1
2x2 + 3x – 20 = 0
oe eg x2 + 1.5x – 10 = 0
A1
(2x – 5)(x + 4) = 0 M1 for an attempt at using an algebraic method such as factorising, formula (allow one error) or completing the square (allow one error) to solve the quadratic eg for (2x + a)(x + b) where ab = ± 20 A1 for a completely correct method
M1dep A1
x = 2.5 Do not award last A1 if a negative value given as possible answer eg if –4 given 2.5 seen with no or incomplete work SC2 2.5 after first M1, A1 give 5/5
A1 [5]
M26. (a) y = x+ 1, y = – x – 2
x + 1= –x – 2 B1
–x – 2 = x2 + 2x – 3
M1
Page 40 of 53
(b) x2 + 3x – 1 = 0
Simplified to 3 terms in x2, x and constant
e.g. x2 = 1 – 3x
A1
x2 + 2x – 3 = × + 1
M1
(c) 1.6, –2.6 Accept 1.5 to 1.6, – 2.5 to – 2.6
A1 [5]
M27. (a) 0 = 4 + 2p + q –5 = 9 – 3p + q
for substitution of both sets of coordinates allow one error
M1
5 = –5 + 5p oe for correct attempt at elimination of p or q
DM1
p = 2 A1
q = –8 p = 2 and q = –8 from no obvious working scores 4
A1
(b) Solving their x2 + px + q = 0
if ‘formula’ used substitution must be completely correct M1
(–4, 0) A1
[6]
M28. (a) Line y= 1 drawn or points on curve
Accept y = 1 written in body of script. M1
0.8, –3.8 (±0.1) A1
Page 41 of 53
(b) Attempt to split equation into
x2 + 3x – 2 = ax + b
Or x2 + 3x – 2 -(x2 + 2x – 1)
Or x2 + 3x – 2 + ax + b = x2 + 2x – 1 M1
Line (y = x – 1) drawn A1
0.4, –2.4 (±0.1) f.t. on their line if Ml awarded, e.g. y = x + 1(1, –3), y = 1 – x(0.6 (0.7), –4.6 (–4.7)),y = –1 –x(0.2, –4.2)
A1 ft [5]
M29. Either, (x + 3)(x – 5) = x2 – 2x – 15
Expansion not necessary for M1 M1
b = –2 A1
c = –15 A1
Note starting with (x – 3)(x + 5) could give c = –15 and will score M1, A0, A1
OR, substituting coordinates (–3,0) and (5,0) into equation to get:
0 = 9 –3b + c and 0 = 5 + 5b + c
correct substitution which might be unsimplified
eg. 0 = (–3)2 – 3b + c and 0 = 52 + 5b + c M1
Solving to give b = –2 A1
c = –15 A1
[3]
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M30. a = 5
from expansion x2 – 2ax + a2 and comparing coeffs.
or simply spotting that a = 5 B1
b = – 7
ft. from their a using a2 + b = 18
ie. b = 18 – a2 or by inspection
B1 ft [2]
M31. (3x + 2)(x + 1)
Rectangle M1
3x2 + 5x + 2
Rectangle A1
x × 3x + 5(x + y) or x × 3x + x × 5 + y × 5 or x(3x + 5) + y × 5 or (3x + 5)(x + y) – 3x × y
M1
3x2 + 5x + 5y
L shape A1
5y = 2 oe dependent on a previous M1 and a term in y
M1 dep
0.4 oe
A1 [6]
M32. (a) (a =) 3 B1
(b =) –12 Allow 12 if –12 given in working
B1
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(b) (x + 3)2 = 12
or (x =)
Using their values from (a) Substitution into formula (allow 1 error)
M1
x + 3 = √12
or (x =)
Using their values from (a) M1 dep
(x =) ±√12 – 3
or A1
[5]
M33. (x – 2) + 5x(x + 1) = 3(x + 1)(x – 2)
Allow 1 error M1
5x2 + 6x – 2 = 3x2 – 3x – 6
A1
2x2 + 9x + 4 = 0
M1
(2x + 1)(x + 4) = 0 A1
x = –1/2, –4 A1
[5]
M34. Sight of 10x or –3(2x – 1) or 3x(2x – 1) M1
–6x + 3 or 6x2 – 3x
M1 dep
6x2 – 7x – 3 (= 0)
A1
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(2x – 3)(3x + 1) (= 0) M1
x = 1.5 or –A1
Full answer with stages clearly shown Strand (ii)
Q1 [6]
M35. 3(x + 2) or 4(x − 1)
Ignore RHS and denominator M1
3x + 6 − 4x + 4
Allow one error Ignore RHS and denominator this also implies first M1 mark
M1
(x − 1)(x + 2)
or x2 + 2x − x − 2 or x2 + x − 2
As denominator M1
2x2 + 4x − 2x − 4 or 2x2 + 2x − 4
Ignore LHS
or 2(x − 1)(x + 2)
this also implies third M1 mark
or 2(x2 + 2x − x − 2) or 2(x2 + x − 2)
M1
2x2 + 3x − 14 = 0
A1
(2x + 7)(x − 2) (= 0)
Correct method to solve their quadratic equation by Correct substitution into the quadratic formula or Correct completion of the square or Correct factorisation
M1
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– and 2
ft their quadratic but must be exact values (integers, fractions, decimals or surds)
SC2 x = − or x = 2
A1ft [7]
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E1. Part (a) was quite well answered. Many students who gave the correct answer in part (a) did not use this to help in part (b) but started again.
E2. This question was a good discriminator, with many students making good progress. Common errors included not dividing the −b by 2a, arithmetical errors and failing to give both answers to two decimal places.
E3. As usual the quadratic formula question proved to be a good discriminator. A significant number of students did not realise that the formula was required with some attempts at trial and error to obtain an answer. Many students appeared concerned that the answers were negative and changed answers to positive as a consequence.
E4. This question was poorly answered. Very few students showed a valid method to find the equation of the straight line. Those who did often incorrectly drew the straight line. Often, when the line was drawn correctly, values for x were not given.
E5. A large proportion of students were unable to access this question suggesting a lack of preparation for this topic. Many of the answers were not given to 2 decimal places.
E6. The algebraic manipulation required in this question proved too challenging for all but the most able students. Many realised that this question required the use of Pythagoras’ theorem but could not expand the brackets correctly. 4x2 + 1, 2x2 + 1 and 4x2 − 4x − 1 were very common errors.
E7. Very few candidates had any success with this challenging question. Many were able to make a start by writing down expressions for the time but few were then able to set up an equation to solve with many resorting to attempts at trial and error to find the solution.
E8. This question was not well answered. Many students used the quadratic formula. Many did not write √40 as 2√10. The students who did know the algorithm for completing the square usually gave a fully correct solution.
E9. This question was a good discriminator, although many did not recognise that the quadratic formula was needed and many failed attempts at factorising were seen. Common errors when using the formula were to use c = 10 or ‒b = + 4. Answers were often given to two decimal places.
E10. A significant majority did not score marks on this standard quadratic formula question with many making no attempt at it. Students who used trial and improvement had no success as they only found one solution at best. A few students attempted to complete the square, usually unsuccessfully. Of those who did realise to use the quadratic formula, the most common errors were using 6 for b or + 5 for c or only having the division line beneath the root.
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E11. This question was a good discriminator. Common errors included just giving the positive square root, not giving answers to 1 decimal place, attempting to solve using trial and improvement and using the quadratic formula but with errors in the values for a, b and c, for example using b = −5.
E12. There were many good attempts at this ‘completing the square’ question. Part (a) assessed Quality of Written Communication and many were unable to complete the algebraic reasoning with sufficient rigour. In part (b), only a few were able to explain that the squared bracket would always be positive or zero and that adding a positive number to it would keep it positive. A large number of students discussed the unrelated issue of odd and even situations.
E13. Although the question directed the students to ‘set up and solve’, fully correct responses were quite rare, with many making no progress at all. Students who did set up a quadratic
equation often reached x2 − x = 110 and then did not how to proceed. Some ignored the 90 and
attempted to solve x2 − x − 20 = 0. Others did not realise that the quadratic would factorise and made errors using the quadratic formula.
E14. This question was generally poorly answered. Some candidates stopped after factorising correctly ie. (5x ‒ 6)(x + 4). Others who factorised correctly then followed up with answers of ‒4 and 6, and did not understand how the answers related to the previous stage. Many rearranged
to 5x + 14x = 24 then unsuccessfully tried to solve this equation. Many of those who attempted
to use the quadratic equation formula did not progress past and those who tried to complete the square were usually unsuccessful.
E15. A small number of candidates realised what to do and usually obtained the correct quadratic equation. Many of these successfully completed the question. Many candidates worked with a and b and did not replace them with x and 3. Teachers should note that a similar question appears in one of the unit 2 practice papers on AQA’s ‘All About Maths’ website.
E16. Many candidates had multiple attempts at this question. The fact that the answer was given in the question should indicate that the fractions need to be cleared. Those who realised this tried various strategies. A common error was to multiply by x + 2 but only for two terms giving
.
2
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E17. A number of students expanded (x ‒ 7) correctly but did not make the connection between ‘ax’ and ‘‒14x’. Some extracted a = ‒14 and b = 49 ‒ (‒14) and then gave b = 35. This error was as common as the correct answer of 63. a = ‒14 and b = 49 was another common error. Some stated that a = 14 after a correct expansion and went on to get the correct corresponding
value of b. Many who tried to expand (x ‒ 7) gave x2 ‒ 49 or x ‒ 14. Of those who obtained four terms, ‒49 (instead of +49) was a common error. A few obtained the correct four terms but then cancelled ‒7x and ‒7x to give x + 49.
E18.
This question is drawn from our specimen paper produced in advance of live examinations. As such, the question was not used in a live examination and therefore no Examiner's Remarks exist.
E19. Most candidates managed at least one mark for correctly using Pythagoras. theorem. The
brackets were then usually expanded correctly with the occasional expected error of (–2)2 = –4. Oddly at this stage, very few continued, to get a quadratic equation. Of those that did, the majority tried factorisation to solve it. The clue to give the answer to an appropriate degree of accuracy was there to imply the need to use the quadratic formula. This form of words was chosen to avoid asking for an answer to 2dp to avoid solutions by Trial and Improvement. Many candidates did try Trial and Improvement, but few managed an answer to sufficient accuracy to gain any marks.
E20. Part (a) was well answered with most candidates being able to expand the brackets. However, it is clear that many candidates have difficulty with factorising. In part (b) candidates often did not realise that the instruction to factorise fully indicated that there was more than one factor. Part (c) was well answered. In part (d) many candidates attempted to factorise and then abandoned this effort and tried to use the quadratic formula. Marks were awarded for their most successful effort. Candidates should be prepared for quadratic equations where the coefficient of x2 is not necessarily a prime number as the specification is not restricted in this way.
E21. Less than 10% of candidates managed to go through all the required steps and obtain a correct solution to this difficult quadratic equation. Many scored one or two marks for correct expansion of the two brackets but often did not know how to proceed from that point. Candidates who appreciated the need to rearrange the equation so that it was equal to zero often made mistakes with the signs. This sometimes meant that candidates found a quadratic equation that gave a square root with a negative number when the quadratic formula was used. Where this occurred it was usually ignored with candidates not appreciating that it meant that a mistake must have been made. About 7% of candidates did not attempt this question.
2
2 2
2
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E22. This question was, in general, poorly done with nearly sixty percent of candidates scoring no marks for this routine piece of algebra. Many candidates attempted to factorise whilst not appreciating that the instruction “Give your answer to 2 decimal places” meant that this was an incorrect approach. There were many errors in the use of the formula most notably the
substitution of b = +5 instead of –5 and in the calculation of (–5)2 which often became –25.
E23. Very few candidates knew how to complete the square. Of those who did most scored 1 mark for b but then made an error with the sign of a.
E24. Part (a) was a lead in to part (c) and is a standard skill. In the main it was well done with the usual errors with signs. Part (b) was a follow through from part (a) and as such was slightly better done than part (a). Part (c) was not well done. The few candidates who did manage to replace the first denominator with the factorised form could not then simplify this. A real lack of understanding of how to add fractions efficiently was apparent. Very few candidates scored full marks on this question.
E25. Overall few candidates scored full marks on this question, but 2 marks for a Trial and Improvement approach was common. Recollection of the area of a trapezium formula was patchy and on substituting values the bracket around (a + b) was often ignored. This was recovered in many cases but not always. Another common approach was to split the shape into a rectangle and a triangle. Poor algebra at this stage prevented further progress for many but pleasingly many found equations of the form x2 + 1.5x = 10 or 2x2 + 3x = 20. A failure to rearrange into the general quadratic form prevented progress for the vast majority but those that did usually went on to solve the equation. Another pleasing aspect was the rejection of the negative root as ‘impossible’.
E26. This is still a topic that causes great difficulty. Very few scored any marks on this question and full marks were very rare. Part (a) was the most successful with about one-half scoring the mark. Parts (b) and (c) were almost always zero marks. Common errors were to expand (x ± 3.3)(x ± 0.3) in part (b) and to use the quadratic formula in part (c). Both of these approaches scored zero as the question states to use the graphs to solve the problems and to show methods clearly.
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E27. This was a question in which the need for candidates to persevere when a solution was not immediately forthcoming was paramount. Most knew to substitute the coordinates of B and C into the equation of the curve but errors were made at this stage (one error was accepted so long as both sets of coordinates were used). Simplifying the equations to get 2p + q = –4 and –3p + q = –14 should not have been a difficult task but far too many candidates gave up too easily. There were a number of fully correct solutions but usually only from strong candidates. Part (b) could not be successfully attempted by those candidates who failed to reach a satisfactory conclusion in part (a). Those who just looked at the diagram and made a guess of (–4, 0) scored nothing. The answer had to be obtained from some legitimate working following through from part (a), albeit allowing for slight error there.
E28. This question was done very badly on the whole with many candidates not even attempting it, though candidates in some centres did much better than others. It seems to be a topic that many candidates have little or no idea how to approach. There are many ways of identifying the required lines but signs and rearranging equations caused problems. A common error in part (a) was y = –1, as was y = – x ± 1 in part (b). Many candidates obtained the answer using the quadratic formula, which gained no credit as it was the wrong method. Correct working needs to be seen in this type of question.
E29. This was a new style of question testing the ‘using and applying’ aspect of the specification, and proved to be too difficult for many. Nonetheless, it is heartening to see candidates who can ‘think on their feet’. The best method was to realise that the given values of –3 and 5 were the solutions of the associated quadratic equation. More popular by far, but also less successful, was to substitute the values of –3 and 5 into the equation of the curve, effectively setting up a pair of simultaneous equations. These ought not to have been too difficult to solve but in practice they were. It is possible (but quite a difficult concept) to obtain a solution by considering the curve as being
a transformation of the curve y = (x – 1)2 (using the fact that x = 1 is a line of symmetry). This solution, although rare, was seen.
E30. Good candidates obtained (x - 5)² – 7 either by inspection or by completing the square but all too often put a = –5 on the answer line. This was penalised despite the presence of the correct expression in the working. Many tried to expand the right hand side but only succeeded in getting some awful algebraic expressions for the answers to a and b. When numerical answers were given they were often just guesses and a significant number of candidates made no attempt at all.
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E31. The techniques required for this question were standard algebraic algorithms. However, the unstructured nature of the question meant that a large number of candidates were unable to identify a valid method for solving the problem. When calculating the area of the rectangle, basic errors such as omitting brackets or writing “(x + 1)(3x + 2) = 3x2 + 2” or
“(x + 1)(3x + 2) = 3x2 + 5x + 3”. When calculating the area of the L-shape, candidates often left out one rectangle.
E32. Centres should be aware that the most successful answers to questions of this type arise from completing the square on the left-hand side rather than expanding the right-hand side and equating coefficients. Having made this comment, very few fully correct answers were seen to part (a). Only a minority of candidates used their answer from part (a) to solve part (b); those who did were generally the more successful. Candidates who used the quadratic formula often had difficulties with signs. A significant number of candidates attempted to factorise, despite the instruction to give the answers in surd form.
E33. Full marks on this question were rare. Many scored the first method mark for combining the numerator of the left hand side correctly but this was often followed by a failure to expand correctly, for example x – 2 + 5x2 + 5 was common. Many gained a method mark for getting the right hand side as 3(x + 1)(x – 2) but this was then expanded incorrectly. If they got this far only
a few knew to rearrange into a quadratic of the form ax2 + bx + c = 0. At this stage earlier errors meant that this was difficult, if not impossible, to solve. This question inevitably was given up on at some stage.
E34.
This question is drawn from our specimen paper produced in advance of live examinations. As such, the question was not used in a live examination and therefore no Examiner's Remarks exist.
E35. This question was not well answered by the majority of students, although it was a good discriminator of the more able students. Some students attempted a trial and error method to obtain a correct answer of 2. Many others eliminated denominators by correctly multiplying and expanding on both sides but poor algebraic manipulation in forming a quadratic equation limited further progress.
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