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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 22:52:59 IST
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16x4 +4x2 +1=0 a,b,c,d are roots find value of a4 + b4 +c4 +d4 1111----------(35) times is this divisible by 3,or 11
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:04:08 IST
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is there no one to solve on goiit
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:07:56 IST
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:10:56 IST
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is answer-1/4
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:13:37 IST
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@sboosy i beleive that question was his signature anyways just complete the squares by adding 4x^{2} to both sides this gives (4x^{2}+1)^{2}=4x^{2} so u have two quadratics 4x^{2}+1-2x=0 roots a,b 4x^{2}+1+2x roots c,d use them to get a^{4}+b^{4} and c^{4}+d^{4} seperately and add them
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KVS
EP@IITM
Godav,IITM |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:14:44 IST
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11111-----35 times is not divisible by both. sum=35 not div by 3 35 is odd,so sum of alternate nos is 17,18 which r not equal.so not div by 11. !!!!!!!!!!!!!!!!!
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Varsha
be cool,
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:15:17 IST
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put 4x=t t^2 + t +1 = 0 t=w, w^2 ,so x= +- (w/2), +-(w^0.5)/2 that gives a^4+b^4+c^4+d^4= 1/8 (w+w^2)= -1/8
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:18:21 IST
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yes they dont have a repeated root I din't know how i made that error. guess i must be sleeeeepy 
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 Mar 2008 23:25:17 IST
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how can the equation have a repeated root-its not a perfect square also in your method RyuAmakusa t=w/4 , w^2 /4
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Apr 2008 10:20:04 IST
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Since a is a root,  . Ditto for b,c, and d. Hence  Since  Hence  or 
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Time wounds all heels |
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