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secondary 3 | A Maths
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Rachel
Rachel

secondary 3 chevron_right A Maths chevron_right Singapore

please help me with (c) and (d) given that a and b are integers :)

Date Posted: 4 years ago
Views: 294
snell
Snell
4 years ago
(c)
(a + b√3)(5 - 3√3)
= 5a - 3a√3 + 5b√3 - 9b
= (5a - 9b) - (3a - 5b)√3

c.f. 6 - 4√3

5a - 9b = 6 ... [1]
3a - 5b = 4 ... [2]

solve simultaneous eqns [1] & [2]

(d)
(2 + 4√7)/(a + b√7)
= (2 + 4√7)/(a + b√7) x (a-b√7)/(a-b√7)
= (2 + 4√7)(a - b√7)/(a² - 7b²)
= (2a + 4a√7 - 2b√7 - 28b)/(a² - 7b²)
= (2a-28b)/(a² - 7b²)
+ [(4a-2b)√7]/(a² - 7b²)

c.f. 4 - √7

(2a-28b)/(a² - 7b²) = 4
(a - 14b)/(a² - 7b²) = 2 ... [1]


(4a-2b)/(a² - 7b²) = -1 ... [2]

solve simultaneous eqns [1] & [2]
Rachel
Rachel
4 years ago
thank you!

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Cheryl Ann Tan
Cheryl Ann Tan's answer
7 answers (Tutor Details)
1st
Hope this helps :)
Eric Nicholas K
Eric Nicholas K
4 years ago
For (d), cross multiplication to interchange the positions is an easier approach.

Edit: actually part (c) same idea.
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Alternative method.
Eric Nicholas K
Eric Nicholas K
4 years ago
You can use this approach because a and b fall under a single sum a + b root something. This method bypasses the need for simultaneous equations.