prove that-

(x^6)+ (x^4) +(x^2)+x+3=0

has no real roots.

there are 2 negative roots..

well this eqn has 2 negatice roots....question is wrong

thats incorrect!

this was simple enough and i figured it out-

all the termsof this equation are to be positive for any real number, because square of any real no. is positive, and the remaining x for sure will have a value which is smaller than (x^6)+(x^4)+(x^2) therefore to get 0, x has to be imaginary.

to get 0.

:D

AJ