Polynomial with odd number of real roots

I have been trying to characterize the number of roots on $\theta$ for the following polynomial

$$\sum_{i=1}^n \frac{\theta- x_i}{1+ \left(x_i - \theta \right)^2} = 0$$

If we were to put everything under a common denominator then we would see that the polynomial is of degree $2n -1$ with coefficients depending on the $x_i$ for $i=1,2,...,n$ . Taking limits as $\theta \to \infty$ and $\theta \to -\infty$ we find that the polynomial has at least one real root as it tends to zero through positive and negative values respectively.

What I do not understand is how one reaches the conclusion that the number of real roots is odd in this case. I know that the complex roots are $2 n -1$ but I'm talking about the real roots. Is there a theorem that guarantees that or is that common sense hiding in plain sight, at least for me?

Thank you.

• What are $x_i$'s? They are real numbers? Commented Feb 27, 2016 at 16:38
• @hamidkamali Yes, they are reals. Commented Feb 27, 2016 at 16:38
• Well, first, I get a polynomial of degree $2n+1$, not $2n-1$. Moreover, if all the $x_i$s are real then the polynomial has of course an even number of complex roots. So there's only an odd number left for the real roots. Commented Feb 27, 2016 at 16:43
• Try with n=2 and you will see 2n-1 is the good. The problem is actually very easy if you discard the expression (which is not a polynomial!) and you consider just the resulting polynomial equation written in general form.The complex roots are necessarily even in their set. Commented Feb 27, 2016 at 17:01

If you write your equation in the form of $p(\theta)=\theta^m+a_{m-1}\theta^{m-1}+...+a_1\theta+a_0=0$, you get that $m=2n+1$ is an odd number and $a_i\in \mathbb R$. So, $p(\theta)$ is a polynomial with real coefficient by odd degree. Thus, $p(\theta)$ has at least one real root. But if $z\in \mathbb C$ is a root of a polynomial with real, then $\overline z$ is to. So the number of complex root of any polynomial with real coefficient, is even. From here, the number of real root of a polynomial with real coefficient, is odd.
• Your welcom. If one of this roots is by repeated order 2. It is mean if $p(x)=a(x-\lambda_1)^2(x-\lambda_2)$ for some $a,\lambda_1,\lambda_2\in \mathbb R$. Commented Feb 27, 2016 at 17:05
• " under what conditions does a third degree polynomial have two real solutions?" if z is a solution, so it $\overline{z}$ so every polynomial will have an even number of of non-real solutions. So if a third degree polynomial has two real solutions, it can have at most one non-real solution which means it must have zero non-real solutions so it must have three real solutions. So to have two real roots it must have one double root (such as $(x-2)^2(3x + 1)$). One can quibble (I would) that a double root is really two roots so this doesn't count. But it's the only possibility. So there.... Commented Feb 28, 2016 at 17:56