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secondary 3 | A Maths
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please help me with (c) and (d) given that a and b are integers :)
Date Posted: 5 years ago
Views: 325
(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]
(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]
thank you!
See 2 Answers
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Hope this helps :)
Date Posted:
5 years ago
For (d), cross multiplication to interchange the positions is an easier approach.
Edit: actually part (c) same idea.
Edit: actually part (c) same idea.
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Alternative method.
Date Posted:
5 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.
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