Page 34 - ACE YR IGCSE A TOP APPR' TO ADD MATH

5. 3 2 = 6 12. 8√3 – √3 + 8√3
1
6x
6x
2
˙
6 = 6 5
6x
2
6x = 2 79√3
1 = 5
x = [2]
3
[2] 4√2 – 3√3
– 1 13. h =
3
6. 4 2a – b = (8 b – 3 ) (√3 – √2) 2
2 4a – 2b = 2 3 – b = 4√2 – 3√3 × 5 + 2√6
4a – 2b = 3 – b 5 – 2√6 5 + 2√6
b = 4a – 3 20√2 + 8√12 – 15√3 – 6√18
= 1 5 – 3a =
2 2 = 4a – 3 25 – 24
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= √3 a = 1 = 20√2 + 16√3 – 15√3 – 18√2
b = 1 = √3 + 2√2
= √(7) – [4] [4]
2
x
x
x
7. (16x )(2 3 − a )(x ) = bx 14. 5(8 · 4) − 2(9 ÷ 9) = 8 x
−4
2a
−2 3 − a
x
x
2
2
4
2a
= 1 8 + 3 (2 )(x )(2 3 − a )(x −6 + 2a ) = bx −4 5(8 · 4)−(8 ) = 2(9 ÷ 9)
2
x 2
7− a 4a − 6
−4
x
1 2 x = bx 8 (20 – 1) = 9 1 2
9
= 2 2 4a – 6 = –4 9 x 171
= a = 0.5 1 2 = 2
8
b = 2 7 – 0.5 [3]
b = 90.51 15. (a) PR = PS + SR
2
2
2
[4] = (√20 + √5) + (√20 – √5)
2
2
1 x PR = √50
2
8. Let 5 = y = 5√2
3y + 5y – 2 = 0 [2]
2
(3y – 1)(y + 2) = 0 (b) tan ∠PRQ = √20 + √5
1
y = or y = –2 √20 – √5
3 20 + 2√100 + 5
1 x 1 1 x =
2
5 = or 5 = –2 20 – 5
2
3 = 3
x = –2 log 3 (rejected) or x = −1.365 [3]
5
[5] √5 + 2 √5 + 2
9. 27 x = 8 4 16. √5 – 2 × √5 + 2
x
˙
7 x 5 + 4√5 + 4
27 x = 4 = 5 – 4
7 8 x = 9 + 4√5
x
˙
27
1 2 x = 4 [3]

56
x = –1.9 17. 1 – √7 – 2(1 + √7 )
[3] (1 + √7 )(1 – √7 )
10. 5 + √6 – 5 + √6 = –3√7 – 1
1 – 7
(5 – √6)(5 + √6)
= 2√6 = 3√7 + 1
25 – 6 6
2√6 1 1
= = √7 +
19 2 6
[2] 1
11. 2√5 – 5√2 × 2√5 – 5√2 a = 6
2√5 + 5√2 2√5 – 5√2 b = 1
= 20 – 20√10 + 50 2 [3]
20 – 50
2
2
= 7 – 2√10 or 2√10 – 7 18. (a) PQ = (2√10) = 40
–3 3 QR = (1 + 3√3) = 28 + 6√3
2
2
2
7
2
2
= – + √10 PR = (√3 – 3) = 12 – 6√3
3 3 [3]
Cambridge IGCSE
TM
174 Ace Your Additional Mathematics
Answers Add Math.indd 174 14/03/2022 12:29 PM