The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
HW3 – Grade B / A
Solving Quadratics by Factorising
1.
y2 + 5y = 0
Solve the equation
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Answer ..................................................................
(Total 3 marks)
2.
(a)
p2 + 7p + 12
Factorise
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Answer ………………………………….........
(2)
(b)
Solve the equation
p2 + 7p + 12 = 0
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Answer ………………………………….........
(1)
(Total 3 marks)
3.
Solve the equation y2 – 4y – 45 =0
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Answer .........................................................................
(Total 3 marks)
4.
(a)
Factorise x + 3x 40
2
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Answer.........................................................................................................................
(2)
(b)
Hence, solve the equation x2 + 3x 40 = 0
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
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Answer x = .................................................................................................................
(1)
(Total 3 marks)
5.
Factorise y2 – 8y + 15
(i)
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Answer ................................................................
(2)
(ii)
Hence solve the equation y2 – 8y + 15 = 0
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Answer ................................................................
(1)
(Total 6 marks)
6.
(a)
Expand and simplify
(x + y)(x – y)
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Answer...............................................................................
(2)
(b)
(i)
Factorise
x2 – 13x + 36
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Answer...............................................................................
(2)
(ii)
Hence, or otherwise, solve the equation x2 – 13x + 36 = 0
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Answer...............................................................................
(1)
(Total 5 marks)
7.
(a)
Solve the equation
1 x5 1 x3
2
4
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
……………...........………………………….....…………………………………….
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Answer x = .................................................................
(3)
(b)
(i)
Factorise
x2 + 5x – 14
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Answer .......................................................................
(2)
(ii)
Hence solve the equation
x2 + 5x – 14 = 0
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Answer .......................................................................
(1)
(Total 6 marks)
8.
The perimeter of a rectangle is 25 cm.
The length of the rectangle is x cm.
x cm
(a)
Not to scale
Write down an expression for the width of the rectangle in terms of x.
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Answer ................................................................... cm
(1)
(b)
The area of the rectangle is 38 cm2.
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
Show that 2x2 – 25x + 76 = 0
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(2)
(c)
Solve the equation given in part (b) to find the value of x.
Give your answer to 2 decimal places.
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Answer .........................................................................
(3)
(Total 6 marks)
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
9.
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.
Not drawn accurately
(x + 1) cm
x cm
(x + 2) cm
Find the value of x.
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Answer x = ......................................................... cm
(Total 5 marks)
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
1.
y(y + 5)
M1
0
A1
Trial & improvement giving 0 or –5 only: SC1
–5
A1
Trial & improvement giving 0 or –5 only: SC1
[3]
2.
(a)
(b)
(p + a)(p + b) where ab = 12
a and b must be positive
M1
(p + 3)(p + 4)
A1
–3 and –4
B1
ft from answer to (a)
[3]
3.
(y  a)(y  b) where ab = 45
M1
(y – 9)(y + 5)
A1
(+)9, –5
A1
ft on their brackets if M1 gained
[3]
4.
(a)
(i)
(ii)
(x  a)(x  b) where ab = 40
B1
(x + 8)(x – 5)
B1
(x =) – 8 and (+) 5
ft their (i) unless correct answer here
B1
[3]
5.
(i)
(ii)
(y ± a)(y ± b) where ab = 15
M1
(y – 3)(y – 5)
A1
(+) 3 and (+) 5
ft from b(i) if M awarded
B1
[6]
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
6.
(a)
x2 – y2
2
B2
B1
2
x + xy – xy – y
(4 terms seen any 3 correct)
(b)
(i)
(x – 9)(x – 4)
B1 (x ± 9)(x ± 4)
(ii)
4 and 9
B2
B1ft
[5]
7.
(a)
(b)
2x – 20 = x + 12
allow one error
or ½ x – ¼ x = 3 + 5
or ¼ x or 8 in ¼ x = 8
M1
allow one error
x – 20 = 12 or 2x = x + 32
or ¼ x = 8
A1
(x =) 32
A1
(x + a)(x + b)
ab = ±14
M1
(x + 7)(x – 2)
A1
–7, 2
B1ft
from two linear factors
[6]
8.
(a)
12.5 – x
M1
oe
(b)
(c)
x(12.5 – x) = 38
M1
Sorting to 2x2 – 25x + 76 = 0
Need valid intermediate step
A1
{25 ± (252 – 4 × 2 × 76)} ÷ 4
M1 allow one error
A1 for correct substitution
7.28 or 5.22 or both
M1 A1
A1
[6]
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
9.
(Area =)
1
x (x + 1 + x + 2)
2
oe (x + 1) +
M1
1
 x  (1)
2
2x2 + 3x – 20 = 0
A1
oe eg
2
x + 1.5x – 10 = 0
(2x – 5)(x + 4) = 0
M1dep
A1
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
x = 2.5
A1
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
[5]
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