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The number of distinct real roots of this equation $$x^9 + x^7 + x^5 + x^3 + x + 1 =0$$ is

Descarte rule of signs doesnt seems to work here as answer is not consistent . in general i would like to know nhow to find number of real roots of any degree equation

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Your polynomial has only odd exponents (apart from the constant term, which causes a shift up the "y-axis"). So in negative x direction it is monotonely decreasing and in positive direction monotonely increasing. All in all you get exactly one real root (because at $x=0$ it has positive value and at $x=-1$ it has negative value.

A second way to solve the problem is taking the derivative (x^8+x^6+x^4+x^2+1) and seeing where this is 0 (never, because all the exponents are positive) So you have neither maximum nor minimum. There is one zero because of the sign change but this must be all (between to zeros there has to be an extremal point).

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Denote the polynomial as $p$.

$p$ have at least one real root because it is of odd degree (this follows from IVT).

Assume $p$ have more then one real root, then there are two different points $x_1,x_2$ s.t $p(x_1)=p(x_2)=0$

By Rolle's theorem this imply there is $x_1<c<x_2$ s.t $p'(x)=0$ - show such a point does not exist

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