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Given xy=1 What is the value of ?

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by Abdulla » Sat Nov 08, 2008 9:31 pm
Given xy=1 What is the value of 2(x+y)^2/ 2(x-y)^2=?
The answer 16..

My answer is :

=2( x^2 + 2xy + y^2 ) / 2( x^2 - 2xy + y^2 )
= x^2 + 2(1) + y^2 / x^2 - 2(1) + y^2

I stopped here.. can someone continue please ?
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Re: Given xy=1 What is the value of ?

by logitech » Sat Nov 08, 2008 9:42 pm
Abdulla wrote:Given xy=1 What is the value of 2(x+y)^2/ 2(x-y)^2=?
The answer 16..

My answer is :

=2( x^2 + 2xy + y^2 ) / 2( x^2 - 2xy + y^2 )
= x^2 + 2(1) + y^2 / x^2 - 2(1) + y^2

I stopped here.. can someone continue please ?
Hint X=1/Y

:idea:
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by Abdulla » Sat Nov 08, 2008 10:00 pm
although I substitute x with 1/y still can't find the answer :(
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by cramya » Sat Nov 08, 2008 10:11 pm
Given xy=1 What is the value of 2(x+y)^2/ 2(x-y)^2=?
The answer 16..

There is a typo in this problem. If I am not mistaken it should read

2 ^ (x+y) ^ 2 / 2 ^(x-y)^2

= 2 ^ (x^2+y^2+2xy) / 2 ( x^2+y^2-2xy)

(a^m / a ^n = a^ (m-n) where a =2 m = x^2+y^2+2xy n = x^2+y^2-2xy)

= 2 ^ (x^2+y^2+2xy-x^2-y^2+2xy)

= 2 ^ 4xy
= 2 ^ 4(1)

=16
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by Abdulla » Sat Nov 08, 2008 10:18 pm
I'm idiot ... Thats why the x & y letters were very small on the screen .. I was asking myself why its too small, but never think its exponents... FUNNY :D

Thank you Cramya
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by Perminology » Thu Dec 24, 2009 12:07 am
Cramya amazing solution! Thanks so much!
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by maaz_gmat » Thu May 13, 2010 6:50 am
Cramya, nice solution. Easy way to be understood by all. Thanks so much.
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by harshavardhanc » Thu May 13, 2010 6:56 am
as the question doesn't have other conditions except XY=1,

take x=y=1, substitute and be happy! :)
Regards,
Harsha
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