Additional Mathematics
11/1/05
Chapter 1: Algebra 1 - Review
Chapter Assessment
1
2
3
(i)
Simplify: a  2b + 2a  3b.
[1]
(ii)
Factorise: 2x2 + 6xy.
[2]
(iii) Simplify:
a 2 b2
 .
b a
[2]
(iv) Simplify:
2m 1
3
 m m.
3 4
5
[2]
Solve the following equations.
(i)
2(x + 5) = 1  x
[2]
(ii)
x 3 1  2 x 

5
2
4
[4]
Make u the subject of the following formula.
v2 = u2 + 2as
4
[2]
3 bottles of wine and 2 packs of beer cost £20.
A bottle of wine costs £5 more than a pack of beer.
Letting x be the cost of a bottle of wine, formulate an equation in x and solve it to find the cost
of a pack of beer and a bottle of wine.
[4]
5
Simplify: (x + 1)2  (x  1)2.
6
Expand: (2x + 3)2.
[3]
Hence solve the equation: (2x + 3)2 = 12x + 109.
© MEI, 2005
[4]
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Additional Mathematics
7
(a)
(b)
8
11/1/05
Solve the following quadratic equations, giving your answers to 3 decimal places where
appropriate.
(i)
x2 + 3x  4 = 0
[3]
(ii)
x2 + 4x  7 = 0
[4]
Express x2 + 6x  19 in the form (x + a)2 + b, where a and b are to be determined.
Hence explain why x2 + 6x  19 = 0 has no real roots.
[5]
Solve algebraically these two simultaneous equations.
2x + 3y + 2 = 0
y = 3x  19
[4]
9
Find the points of intersection of the line y + 3x + 3 = 0 and the curve y = x2  5x  3.
10
The line y + 4x = k is a tangent of the circle x2 + y2 = 17.
[7]
(i)
Make y the subject of the equation y + 4x = k.
(ii)
Substitute into the equation for the circle to obtain a quadratic equation in x involving k.
[4]
(iii) Write down the condition for this equation to have coincident roots.
[1]
[3]
(iv) Solve this quadratic equation in k to give the two values of k and hence the equations of
the two possible tangents.
[3]
Total: 60
© MEI, 2005
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