Question

(x+3)^4+(x+5)^4=16

Answer 1

x=-3

Answer 2

i wonder

say r = x+3

(r)^4+(r+2)^4=16

Answer 3

and the rest of 3 solution are

x = -5

Answer 4

we have to find the no.of real roots

Answer 5

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

Answer 6

ida prolly determined r=0; therefore x=r-3, x=-3

Answer 7

@experi. ur anwer is right .........now can u tell me brief explaintion....i meant step by step

Answer 8

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.

Answer 9

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

Answer 10

O my frnd Experiment..........thnx uuuuu soooooo muchhhhh.....i have got it

Answer 11

Hehehhehe. Everyone, I have posted this question two times before. Payal didi copied me?????????????????????

Answer 12

hapy?

Answer 13

I was already happy much earlier. Anyway, Thanks Arvind Bhai.