Summer Mathematics C1 Transition Answers
1.
Write in index notation
a) 6 4 x 6 2 = 66 b) (53 ) 4 = 512 c) 8 7 ÷ 8 3 = 84 d) 5 2 ÷ 5 4 = 52 e) 4 3 x 4 3 = 40 f) 4 3 x 2 8 = 214
2.
Find the value of
a) 10
3.
1
1
1
= 0.1= 10 b) 2 5 = 32
1
4
c) 81 = 3
1
=5
d)
7 x
200 -
180
= √36 = 6 c)
75 = 5√3
5
72 + 18 = 10√2 − 6√2 + 3√2 = 7√2
7 =7
b)
Rationalize and simplify as appropriate
5.
Make “a” the subject of the formula in each case
a) 4a + 12b = 32c
b) 3d =
a  10
a)
1
4.
3
=
√3
3
c) 2f = 5a 2 - 11q
𝑎 + 3𝑏 = 8𝑐
9𝑑2 = 𝑎 − 10
2𝑓 + 11𝑞 = 5𝑎2
𝑎 = 8𝑐 − 3𝑏
𝑎 = 9𝑑2 + 10
√
2𝑓+11𝑞
5
5
b)
8.
9.
√5
= 5√3=
75
√5√3 √15
= 15
15
d) -2a +14 = 6ba + 14t
14 − 14𝑡 = 𝑎(6𝑏 + 2)
𝑎=
=a
14−14𝑡
6𝑏+2
7−7𝑡
=3𝑏+1
Expand the following simplifying the answer where possible
a) (x – 3)(x + 4)
b) (2x – 7)(x + 8)
c) (3x +2)(3x -2)
=𝑥 2 + 𝑥 − 12
= 2𝑥 2 + 9𝑥 − 56
=9𝑥 2 − 4
e) Simplify ( 5  2 ) 2 leaving your answer in the form a + b c .
= 5 + 4√5 + 4 = 9 + 4√5
7.
2
1 3
e) 49 = 343 f)   = 4
8
3
2
Simplify these surds
a)
6.
d) 25
1
2
d) (5x + 13) 2
=25𝑥 2 + 130𝑥 + 169
Factorise each of these quadratic expressions
a) x 2 + 3x + 2
b) x 2 + 20 - 9x
c) x 2 – 49
= (𝑥 + 2)(𝑥 + 1)
d) 49 – x 2
= (7 − 𝑥)(7 + 𝑥)
= (𝑥 + 7)(𝑥 − 7)
f) -5x – 6 + 6x 2
= (2𝑥 − 3)(3𝑥 − 2)
= (𝑥 − 4)(𝑥 − 5)
e) 2x 2 + 7x + 3
= (2𝑥 + 1)(𝑥 + 3)
a) x 2 – x – 6 = 0
(𝑥 − 3)(𝑥 + 2) = 0
𝑥 = 3 𝑥 = −2
b) 3x 2 + 5x = 2
3𝑥 2 + 5𝑥 − 2 = 0
(3𝑥 − 1)(𝑥 + 2) = 0
1
𝑥 = 3 𝑥 = −2
Some quadratic expressions do not factorize. You may be asked to complete the square.
Solve these quadratic equations
a) Show that ( x + 3 ) 2 - 8  x 2 + 6x +1
(𝑥 + 3)2 − 8 = 𝑥 2 + 6𝑥 + 9 − 8 = 𝑥 2 + 6𝑥 + 1
b) Write x 2 + 4x – 3 in the form ( x + a ) 2 + b 𝑥 2 + 4𝑥 − 3 = (𝑥 + 2)2 − 4 − 3 = (𝑥 + 2)2 − 7
c) Hence, solve x 2 + 4x – 3 = 0 leaving your answer in surd form.
(𝑥 + 2)2 − 7 = 0
𝑥 + 2 = ±√7
𝑥 = −2 ± √7
10.
Solve these equations simultaneously (remember to find values for x and y)
by elimination…
a) 4x + y =10
2x – y = 8
by substitution…
b) 4x – 5y = 4
2y = 25 – 6x
2𝑥 − (10 − 4𝑥) = 8
𝑥 = 3 𝑦 = −2
12𝑥 = 50 − 4𝑦
12 + 15𝑦 = 50 − 4𝑦
𝑦 = 2 𝑥 = 3.5
c) x + y = 11
xy = 30
𝑥(11 − 𝑥) = 30
0 = 𝑥 2 − 11𝑥 + 30
𝑥=5 𝑥=6
𝑦=6 𝑦=5
Substitution above!
Elimination below.
a) 4x + y =10
2x – y = 8
6𝑥 = 18
𝑥 = 3 𝑦 = −2
b) 4x – 5y = 4
2y = 25 – 6x
12𝑥 = 50 − 4𝑦
12𝑥 = 12 + 15𝑦
0 = 38 − 19𝑦
𝑦 = 2 𝑥 = 3.5
d) 3x + y = 14
x + y 2 = 28
𝑥 + (14 − 3𝑥)2 = 28
9𝑥 2 − 83𝑥 + 168 = 0
56
𝑥=3 𝑥= 9
𝑦=5
𝑦=−
14
3