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sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =

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Manager
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sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) = [#permalink] New post 13 Sep 2005, 12:36
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(sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) = ?
VP
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 [#permalink] New post 13 Sep 2005, 13:14
(sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3))

multiple both numerator and denominator by :
(sqrt(5)+sqrt(3))

gives:

{[(sqrt(5)+sqrt(3))] ^ 2} / 2

or

4 + sqrt (15)
Manager
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 [#permalink] New post 13 Sep 2005, 19:34
the reason i posted this question was to ask a second question.

if you chose to multiply the numerator and denominator by sqrt(5)+sqrt(3) you get 4+sqrt(5)*sqrt(3)

though, if you chose to multiply the numerator and denominator by sqrt(5)-sqrt(3) you get 1 / (4-sqrt(5)*sqrt(3))

though seemingly different,

4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3))

I am not sure whether there is a broad generalization i am missing. If anyone can shed light on this, that would be great!
Senior Manager
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Re: Easy PS: [#permalink] New post 13 Sep 2005, 22:56
chets wrote:
(sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) = ?


Take conjugate on both sides. which gives.

(sqrt(5) + sqrt(3)) (sqrt(5) + sqrt(3)) / (sqrt(5) - sqrt(3)) (sqrt(5) + sqrt(3))

This gives (sqrt(5) +sqrt(3))^2 / (Sqrt(5)^2) - (sqrt(3))^2

=> 5 + 3 + 2 sqrt(15) / 2

=> 2(4 + sqrt(15))/2

=> 4 + sqrt(15)

Thanks
Senior Manager
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 [#permalink] New post 14 Sep 2005, 05:34
chets wrote:
the reason i posted this question was to ask a second question.

if you chose to multiply the numerator and denominator by sqrt(5)+sqrt(3) you get 4+sqrt(5)*sqrt(3)

though, if you chose to multiply the numerator and denominator by sqrt(5)-sqrt(3) you get 1 / (4-sqrt(5)*sqrt(3))

though seemingly different,

4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3))

I am not sure whether there is a broad generalization i am missing. If anyone can shed light on this, that would be great!



Chet,

You r not missing anything,

You get:

1/(4-sqrt(5)*sqrt(3)) =


(4+sqrt(5)*sqrt(3)) / [(4-sqrt(5)*sqrt(3))(4+sqrt(5)*sqrt(3))] =


(4+sqrt(5)*sqrt(3)) / (16 - 15) =


(4+sqrt(5)*sqrt(3))

Hope that helps :)
  [#permalink] 14 Sep 2005, 05:34
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