IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

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

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

akhil_o

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

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

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

sandeepramesh

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

sandeepramesh

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