1. Quadratic Equations and Expressions
New section. This looks at a topic by grade tariff of increasing difficulty. 1
©AQA GCSEMathsB/Spr2004

2. x(x+5) B1 Grade D 1. Factorise x2 + 5x
Answer (1 mark)
x(x+5) B1
Only one mark so right or wrong.
Note that this could be on a Foundation paper.
Foundation candidates do not need to know the term quadratic.
Grade D 2
©AQA GCSEMathsB/Spr2004

3. 2. Complete the table of values for y = 2x2 – 2x + 1.
(2 marks)
(b) Use the table to draw the graph of y = 2x2 – 2x + 1
on the grid below 13 1 y 4 3 2 -1 -2 x B2
–1 eeoo 5 5 25
B1ft Accurate plotting
B Smooth curve (within tolerance)
-1 eeoo means minus one each error or omission.
We expect candidates to be able to plot points and draw a curve. Joining points with a ruler will be penalised.
Grade C 3
©AQA GCSEMathsB/Spr2004

4. 3. Expand and simplify
(a) (x + 3)(x – 4)
Answer (2 marks)
(b) (2x – 3y)(5x + 2y)
Answer (3 marks)
x2 + 3x - 4x - 12
M1 - allow one error
x2 - x - 12 A1
M1 - allow one error
A1 - if all correct
10X2 - 15xy + 4xy - 6y2
10X2 - 11xy - 6y2
A1ft if M1 awarded
Expanding quadratics requires a square term.
Part (b) can be followed through if one error made.
Grade B 4
©AQA GCSEMathsB/Spr2004

5. (x + 5)(x - 5) Grade B 4. Factorise x2 - 25
Answer (1 mark)
(x + 5)(x - 5) B1
Right or wrong
Grade B 5
©AQA GCSEMathsB/Spr2004

6. (x – 2)(x + 7) x = 2 or – 7 Grade B 5. (a) Factorise x2 + 5x – 14
Answer (2 marks)
(b) Hence, or otherwise, solve the equation x2 + 5x – 14 = 0
Answer (1 mark)
M1 if (x ±a)(x ± b) where ab=14
(x – 2)(x + 7)
A1 - if correct
B1ft their brackets if M1 awarded
x = 2 or – 7
Alternative version:
Solve the equation x2 + 5x – 14 = 0
(3 marks)
Method marks awarded for showing some knowledge of how to factorise.
If brackets wrong the last mark can still be awarded
Grade B 6
©AQA GCSEMathsB/Spr2004

7. 4(x + 1) + 2x = 5x(x + 1) 6x + 4= 5x2 + 5x 5x2 – x – 4 = 0
6. (a) Show that can be written as 5x2 – x – 4 = 0.
(2 marks)
(b) Solve the equation 5x2 – x – 4 = 0.
Answer (3 marks)
4(x + 1) + 2x = 5x(x + 1) M1
6x + 4= 5x2 + 5x
5x2 – x – 4 = 0 A1
(5x + 4)(x - 1)= 0
M1 A1
x = -4/5 or 1
As the answer is given it is enough to get to the second line.
We can then assume that they can get to the answer.
They will fiddle it anyway.
Part (b) as before A1
Grade A 7
©AQA GCSEMathsB/Spr2004

8. (2x - 3)(x + 2) = 22 2x2 - 3x + 4x - 6 – 22 = 0 2x2 + x - 28 = 0
7. (a) The rectangle has an area of 22 cm2.
Show that x is a solution of the equation 2x2 + x – 28 = 0
(3 marks)
(b) Find the value of x.
Answer (3 marks)
2x - 3
x + 2
Not to scale
(2x - 3)(x + 2) = 22 M1
2x2 - 3x + 4x - 6 – 22 = 0 A1
2x2 + x - 28 = 0 A1
(2x - 7)(x + 4)= 0
M1 A1
Method marks awarded for setting up the equation.
Accuracy for expanding and partly rearranging.
Part (b) as before
x = 31/2 A1
Grade A 8
©AQA GCSEMathsB/Spr2004

9. x= -4 ± (-4)2 - 4(1)(-6) 2(1) x = -5.16 or 1.16 (x + 2)2 - 10 =0
8. Solve the equation x2 + 4x – 6 = 0
Give your answer to 3 significant figures (paper 2 calculator)
Answer (3 marks)
x= -4 ± (-4)2 - 4(1)(-6)
M1 A1
2(1)
x = or 1.16 A1
Or: Give your answer in surd form (paper 1 non-calculator)
(x + 2) =0
x = -2 ± 10
M1 A1 A1
(3 marks)
Students should be encouraged to substitute into the formula and not work things out.
Then evaluate square root.
Completing the square is a method should be known.
Grade A 9
©AQA GCSEMathsB/Spr2004

10. 9. The sketch shows the circle x2 + y2 = 65 and the line x + 2y = 10.
The circle and the line intersect at A and B
(a) Show that the y-coordinates of A and B satisfy the equation y2 – 8y + 7 = 0
(3 marks)
(b) Hence find the coordinates of A and B.
Answer A(……… , …….. )
B(……… , …….. ) (3 marks) x y O B A
Not to scale
(10 - 2y)2 + y2 = 65 M1
5y2 - 40y = 0 A1
y2 - 8y + 7 = 0 A1
(y - 1)(y - 7)= 0 M1
y = 7 or 1 A1
Method for substituting.
Accuracy for expanding.
Accuracy for dividing by 5.
Part (b) as before. A1
Grade A 10
©AQA GCSEMathsB/Spr2004

11. (a - 5b)(a + 5b) 6(a2 - 16b2) 6(a - 4b)(a + 4b) Grade A 10. Factorise
(a) a2 – 25b2
Answer (2 marks)
(b) 6a2 – 96b2
Answer (3 marks)
Allow M1 if (a +xb)(a - xb) where x = 12.5 or 25
A1 - if correct
(a - 5b)(a + 5b)
6(a2 - 16b2) B1
6(a - 4b)(a + 4b)
Method for knowing the form of DOTS.
Part (b) is an alternative scheme. B1 B1
Grade A 11
©AQA GCSEMathsB/Spr2004

12. 4(x + 2) - (x + 3)=(x + 3)(x + 2) x2 + 2x + 1 = 0 x = -1 Grade A*
11. Solve
Answer (5 marks)
4(x + 2) - (x + 3)=(x + 3)(x + 2) M1 M1 M1
x2 + 2x + 1 = 0 A1
x = -1 A1
Three method marks for the first line.
First is for getting the numerator on left correct.
Second for either denominator or RHS.
Third for putting them equal.
Grade A* 12
©AQA GCSEMathsB/Spr2004

13. (x + 3)2 - 9 - 3 a = 3, b = 12 (-3, -12) Grade A*
12. (a) Find the values of a and b such that
x2 + 6x – 3 = (x + a)2 – b
Answer (3 marks)
(b) Hence, or otherwise, write down the minimum point of
y = x2 + 6x – 3
Answer (1 mark)
(x + 3)
M1 A1
a = 3, b = 12 A1
(-3, -12) B1
Typical completing the square scheme.
Last part needs a knowledge of the connection between equation and graph.
Grade A* 13
©AQA GCSEMathsB/Spr2004