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

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sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 13 Sep 2005, 12:36
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\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =\)
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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 22 May 2014, 18:37
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Multiply numerator & denominator by \((\sqrt{5} + \sqrt{3})\)

\(= \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}\)

\(= \frac{5 + 2\sqrt{15} + 3}{5 - 3}\)

\(= \frac{8 + 2\sqrt{15}}{2}\)

\(= 4 + \sqrt{15}\)

Formulae used:

\((a+b)^2 = a^2 + 2ab + b^2\)

\((a-b)^2 = a^2 - b^2\)
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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)
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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!
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Re: Easy PS:  [#permalink]

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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
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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 :)
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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 21 May 2014, 07:38
This type of questions can be solved by using formula

(a+b)(a-b) = a^2 - b^2

Denominator is in the form of a-b , hence we will multiply numerator and denominator by the form a +b .i.e. sqrt(5) + sqrt(3)

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

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

5+3 + 2*sqrt(5)*sqrt(3) / 5-3

8+2*sqrt(5)*sqrt(3)/2

4+sqrt(15)
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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 22 May 2014, 02:33
chets wrote:
\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =\)


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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 22 May 2014, 13:59
Chets asked if 4+sqrt(5)*sqrt(3) = 1/(4-sqrt(5)*sqrt(3)) can be extrapolated into a broad generalization. No, it can't be!

Here's why.

Consider an expression \(1/(a-\sqrt{bc})\) [modeled on 1/(4-sqrt(5)*sqrt(3)) ]
Multiplying Numerator and Denominator with \((a+ \sqrt{bc})\), we get
\((a+ \sqrt{bc})/(a^2-bc)\)

So, the reason why 1/(4-sqrt(5)*sqrt(3) got simplified into 4+sqrt(5)*sqrt(3) was because the Denominator of the above expression is equal to 1 in this particular case (a=4, b=5, c=3. So,\(a^2-bc = 4^2-5*3 = 1\))
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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =  [#permalink]

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New post 23 Sep 2019, 15:23
chets wrote:
\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} =\)


the answer is \(4 + \sqrt{15}\)
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Re: sqrt(5)+sqrt(3)) / (sqrt(5)-sqrt(3)) =   [#permalink] 23 Sep 2019, 15:23

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

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