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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: solve this one if you are good in quadratics by shortest methood
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ziauddin75 (307)

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16x4  +4x2  +1=0
a,b,c,d are roots
find value of    a4 + b4 +c4 +d
 
 
1111----------(35)   times   
is this divisible by 3,or  11
    

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ziauddin75 (307)

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is there no one to solve on goiit
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sboosy (3065)

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1111....(35)\ \mbox{times is not divisible by both 3 and 11} \\ \\ \mbox{Sum of digits is 35.Hence not divisible by 3} \\ \\ \mbox{It is not  divisible by 11 as odd number of 1
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ankitagg (330)

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is answer-1/4

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pardesi (580)

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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

KVS
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varshavallig (800)

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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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tanmoyoutthere (18)

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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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RyuAmakusa (942)

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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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tanmoyoutthere (18)

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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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hsbhatt (6235)

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Since a is a root,
 
16a^4+4a^2+1 = 0. Ditto for b,c, and d.
 
Hence 16sum a^4+4sum a^2+4 = 0
 
Since sum a = 0
 
sum a^2 = -2sum ab = -rac{1} {2}
 
Hence 16sum a^4 = -4 (1+sum a^2) = -2
 
or sum a^4 = -rac{1}{8}

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