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Algebra

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3 May 2008 13:41:12 IST

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Two Variable Equation

None

Find $x, y \in \mathbb{R}$ such that

$x^2+20x+351 = \frac{2008}{y^2+30y+233}$

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Comments (3)

Akhil

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Joined: 17 Jan 2008

Posts: 1451

3 May 2008 14:48:07 IST

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$x^2+20x+351 \,\, minima\, is \,at\,x=-10\\minimum\,value\,of\,LHS=251$

$y^2+30y+233\\minima\,is\,at\,y=-15\\minimum\,value \,is\,8\\Maximum \,value\,of\,RHS\,=2008/8=251$

$hence \,LHS>=251,\\RHS<=251\\Equality\, can\, only\, hold\, at\, x=-10,y=-15$

sandeep ramesh

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Joined: 13 Mar 2008

Posts: 1183

3 May 2008 15:13:09 IST

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dunno if you filched it from there sir or vice versa but here's the same problem :D :D

http://iit-redefined.theforum.name/viewtopic.php?f=14&t=367

sandeep ramesh

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Joined: 13 Mar 2008

Posts: 1183

3 May 2008 15:14:45 IST

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judging by the times in which this ques was posted, gaurav filched it :D :D

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