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i wonder say r = x+3 (r)^4+(r+2)^4=16
and the rest of 3 solution are x = -5
we have to find the no.of real roots
and we have two complex solution, x = -4-i * sqrt(7) and -4 + i * sqrt(7) so in total we have 4 solutions for fourth degree equation two real and two complex real solutions would be -3 and -5
ida prolly determined r=0; therefore x=r-3, x=-3
@experi. ur anwer is right .........now can u tell me brief explaintion....i meant step by step
the best approach to find the real roots would be to see question carefully, (x+3)^4+(x+5)^4 = 2^4 (if x+3 and x+5 were both +ve integers it would be violation of Fermat's theorem) the equation is satisfies if one of them is 2 and other is zero ... so -3 and -5 must the answer.
the other way is to expand and simplify, and the way to solve a quartic equation is given here http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations
O my frnd Experiment..........thnx uuuuu soooooo muchhhhh.....i have got it
Hehehhehe. Everyone, I have posted this question two times before. Payal didi copied me?????????????????????
I was already happy much earlier. Anyway, Thanks Arvind Bhai.