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try substituting each points , the zeros of the equation is actually the solution of the equation
and x^3 being present do you think what i am thinking? :)
haha @hartnn great minds think alike :)
lol .
@hartnn you fell into the trap! read the problem statement again
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O.o there can be 2 complex roots....
REAL gosh !
i go with my first reply :)
and there is!
@AravindG I tried your method and I got d?
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you are right !
This problem illustrates the importance of reading carefully, instead of seeing what you expect...
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