The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
HW4 – Grade A / A*
Solving Quadratics by the Formula and Completing the Square
1.
The base of a triangle is 7 cm longer than its height.
The area of the triangle is 32 cm2.
(a)
Taking the height to be h cm, show that
h2+ 7h  64 = 0
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(3)
(b)
Solve this equation to find the height of the triangle.
Give your answer to 2 decimal places.
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Answer....................................................................................................................cm
(3)
(Total 6 marks)
2.
Solve the equation
x2 + 4x – 10 = 0
Give your answers to 2 decimal places.
You must show your working.
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Answer .................................................................
(Total 3 marks)
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
3.
Solve the equation
x2 – 10x – 5 = 0
Give your answers to 2 decimal places.
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Answer ............................................
(Total 3 marks)
4.
Solve the equation
x2 – 2x – 5 = 0
giving your answers to 3 significant figures.
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Answer …...................................................................
(Total 3 marks)
5.
(a)
Find the values of a and b such that
x2 + 10x + 40 = (x + a)2 + b
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Answer a = ......................... , b = ...............................
(2)
The Robert Smyth School
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
(b)
Hence, or otherwise, write down the minimum value of x2 + 10x + 40
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Answer .........................................................................
(1)
(Total 3 marks)
6.
(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)
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
7.
(a)
Find the values of a and b such that
x2 + 8x – 5 ≡ (x + a)2 + b
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Answer a = ......................... b = ...........................
(2)
(b)
Hence, or otherwise, solve the equation
x2 + 8x – 5 = 0
Give your answers to 2 decimal places.
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Answer .......................................................................
(2)
(Total 4 marks)
The Robert Smyth School
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
1.
(a)
h + 7 seen
M1
½h(h + 7) = 32 oe
M1
sorted to h2 + 7h –64 = 0
(b)
{ – 7  72 – 4x – 64}/2
M1
 305 or 17.46 ... seen
A1
5.23
A1
[6]
2.
Substituting in formula:
– 4  (4) 2 – 4  1  –10
2
Allow one error
Errors are: wrong sign for b (+4)
b2 wrong (=8 or –16)
– 4ac = – 40
M0 for any of the following:
Not dividing whole of top line by 2a
Forgetting square root
Mis-copying formula
– 5.74
M1
A1
Note: some working must be seen answers without working
scores M0A0A0
1.7 and – 5.7 or 1.741 and – 5.741
M1 A1 A0
1.74
A1
Alternative method
(completing the square):
(x + 2)2 – 14 = 0
x = – 2  14
x = – 5.74, 1.74 (both)
M1
A1
A1
[3]
3.
(x – 5)2 – 30 = 0
M1
2
For attempt at (x – 5) ,
x = 30 + 5
A1
For –5 and –30
x = 10.48, –0.48
A1
Both answers (Accept 10.5, –0.477)
ALTERNATIVE
x=
10  10 2  4  1  5
2 1
For substitution into formula (allow one error)
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M1
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The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
x=
10  120
2
A1
Correct substitution
x = 10.48, –0.48
A1
Both answers (accept 10.5, –0.477)
[3]
4.
(x – 1)2 – 6 = 0
M1
Use of formula:
2  4 – 4  1  –5
x=
21
(allow 1 error in formula from wrong
sign for b (–2), b2 as – 4, 4ac as – 4 × 1
× 5 = – 20)
Not dividing whole top by 2a is M0
unless recovered.
x = 1 ± 6
A1
x=
2  24
2
x = – 1.45, 3.45
A1 f.t.
f.t. their a and b or their formula if
one error and root is not negative.
i.e. wrong sign for b (1.45, –3.45)
b2 as – 4 (3, – 1)
If CTS used and 1 +  6 = 3.45 only
answer give M1, A1, A0.
[3]
5.
(a)
(a =) 5
B1
(b =)15
(b)
B1
15
B1 ft
their b
[3]
6.
(a)
(a =) 3
B1
(b =) –12
B1
Allow 12 if –12 given in working
(b)
(x + 3)2 = 12
or (x =)
– 6  6 2 – 4(1)(–3)
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Using their values from (a)
Substitution into formula (allow 1 error)
M1
6
The Robert Smyth School
Mathematics Faculty
Quadratics
Innovation & excellence
x + 3 = 12
– 6  36  12
or (x =)
2
Using their values from (a)
M1 dep
(x =)
±12 – 3
or
– 6  48
2
A1
[5]
7.
(a)
(a)
(x + 4)2
M1
a = 4, b = –21
Do not award if b given as 21
A1
Alt. = x2 + 2ax + a2 + b and
8 = 2a or –5 = a2 + b
M1
a = 4, b = –21
(b)
A1
x = ±√21 ± 4 or x = ±√21 – 4
ft their ‘a’ if solvable for M1 and A1
T&I M0 unless both answers given.
x = 0.58, –8.58
Accept 0.583
x = √21 – 4 = 0.58 SC1
Allow SC1 on follow through for positive root only.
(b)
Alt. x =
–8
M1
A1ft
(8 2 – 4  (1)  –5
Allow x=
2
–8
x = 0.58, –8.58
Accept 0.583
(64 – 20)
2
M1
as only error for M1
A1
[8]
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