Solving Quadratic Equation149
Solving Quadratic Equation149
Solving Quadratic Equation149
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)
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Q2.
I am thinking of two numbers.
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(1)
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or
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Q3.
The graph of y = x2 + 2x − 3 is drawn below.
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)
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Calculator
Q4.
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b = ______________________________________
(Total 3 marks)
Q5.
The diagram shows a rectangle.
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x = ___________________________________ cm
(Total 5 marks)
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Q6.
(a) Expand and simplify (x + 5)(x − 4)
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Answer _________________________________________
(2)
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Answer _________________________________________
(1)
(Total 3 marks)
Q7.
(a) Factorise x2 − 9x + 20
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Answer _________________________________________
(2)
(b) Solve x2 − 9x + 20 = 0
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Answer _________________________________________
(1)
(Total 3 marks)
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Q8.
The area of this rectangle is 28 cm2
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Answer ______________________________________ cm
(Total 7 marks)
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Q9.
(a) Simplify fully 5x2 × 3y4 × 2x × y3
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Answer _________________________________________
(2)
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(2)
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Answer _________________________________________
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Answer _________________________________________
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(Total 7 marks)
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Solutions | Solving Quadratic Equations|Maths:149
Mark schemes
Q1.
(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]
Q2.
(a) x + 7.5 or 7.5 + x
x+7
B1
x2 + 7.5x = 4x + 15
M1
x2 + 3.5x − 15 = 0
or
2x2 + 7x − 30 = 0
M1
(2x − 5)(x + 6) (= 0)
M1
2.5 and 10
either order but in correct pairs
and
−6 and 1.5
SC1 one correct pair
A1
[6]
Q3.
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(x2 + 2x − 3) − (x2 + x − 3)
Or attempt to ‘balance’ equations
M1
y=x
A1
Q4.
(x − 3)(x + 3)
Substitutes any value for x into both
expressions but not x = 0
M1
(x − 3)(x + 5)
Sets up a correct equation in b
M1dep
(b =) 2 or x 2 + 2x − 15
A1
[3]
Q5.
(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
(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]
Q6.
(a) x2− 4x + 5x − 20
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Allow one error
M1
x2 + x − 20
A1
(b) 8 and −7
B1
[3]
Q7.
(a) (x − 4)(x − 5)
B1 for (x − a)(x − b) where ab = 20
or a + b = −9
B2
(b) 4 and 5
ft their part (a) provided two brackets
B1ft
[3]
Q8.
(x + 2)(6x – 1) = 28
M1
6x2 – x + 12x – 2 = 28
Allow one error
M1dep
6x2 +11x – 30 (= 0)
Collect terms to one side, ft their four terms
M1dep
(3x + 10)(2x – 3) (= 0)
A1
(x = – and) x = 1.5
oe
ft their two brackets
B1ft
or 14 × 1.5 + 2
2(6x – 1 + x + 2)
or 14x + 2
M1
23
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(and x = – discarded)
May be implied
B1ft
[7]
Q9.
(a) 30x3y7
B1 for two correct terms
B2
Additional Guidance
30 × x3 × y7
B1
30 × x3y7
B1
x3y730
B1
7x3 × 4y7
B1
eg 10x3 + 3y7
B0
(b) x2 – 3x + 7x – 21
Allow one error
M1
x2 + 4x – 21
A1
Additional Guidance
x2 – 3x – 21 or x2 + 7x – 21 (one error)
M1A0
x2 – 21 (two errors)
M0A0
(c) 8 and –2
or x = 8 and x = –2
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Any order
B1
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