Just a comment, not an answer, but perhaps worth a try.
Assuming $x \not= -\alpha_1,\ldots,-\alpha_n$, we can rewrite the equation as \begin{equation*} \prod_j (1+x\alpha_j)\sum_i \frac{(1-x^2\alpha_i)}{1+x\alpha_i} = 0. \end{equation*} Under our assumption, this further simplifies to the OP's older equation: \begin{equation*} f(x) := \sum_i \frac{1-x^2\alpha_i}{1+x\alpha_i}=0. \end{equation*} Perhaps at this point, you could try to: (i) bracket the root, obtaining points $a<x^*<b$ such that $f(a)<0$, $f(x^*)=0$ and $f(b)>0$, where the bounds are reasonable functions of the $\alpha_i$; (ii) run (by hand) an iteration or two of Newton's method to get an approximate $x^*$ (seems hard); or (iii) try Lagrange inversion.