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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
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
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