tag:blogger.com,1999:blog-3598076976656815157.post5504719223920137310..comments2011-09-27T14:39:12.997+05:30Comments on RamanuGems...! :): RamanuGem_004_The Innocuous(looking!) pair of Equations...! :)Calvinhttp://www.blogger.com/profile/16702462945374110935noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-3598076976656815157.post-84335570451048993232011-09-27T14:34:02.934+05:302011-09-27T14:34:02.934+05:30I am a moron in Maths. I feel amazed when I see su...I am a moron in Maths. I feel amazed when I see such problems.sunaathhttps://www.blogger.com/profile/13386371953472087631noreply@blogger.comtag:blogger.com,1999:blog-3598076976656815157.post-33609729447350629322010-12-27T23:43:47.932+05:302010-12-27T23:43:47.932+05:30looking at the expression one can easily deduce th...looking at the expression one can easily deduce that x , y >0.<br /><br />With that in place the second deduction wud be x,y would be perfect squares <br /><br />With this try doing hit & trial & one wud end up with one resultTarunhttps://www.blogger.com/profile/03076867354115530202noreply@blogger.comtag:blogger.com,1999:blog-3598076976656815157.post-86154470758429692632010-12-27T21:37:13.830+05:302010-12-27T21:37:13.830+05:30Anyway, the "normal", "formal"...Anyway, the "normal", "formal" {But something i HATE...becoz it's NOT b'ful...! Too HARD..!! :-/ } way to solve this baby, is to see that,<br />y = (11-x)^2,<br />& to put that into the equation :<br />x = (7-y)^2<br />=> x - ( 7-(11-x)^2) ^2 = 0<br />x^4 - 44x^3 + 712x^2 - 5017x + 12996 = 0<br />This, friends, is the correct equation to solve!<br />And luckily (x-9), is one of the factors..! :) i.e., the above equation can be factorized as :<br />(x-9)(x^3 - 35x^2 + 397x - 1444) = 0<br />Anyway, me discussing with some friends about another ridiculous way I've devised...! ;)<br />Will update you once/if that's confirmed..! :)Calvinhttps://www.blogger.com/profile/16702462945374110935noreply@blogger.com