Let $E_n(z)$ be the Eulerian polynomial
$$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$
where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$.
These are studied in great detail, see its [OEIS wiki page](http://oeis.org/wiki/Eulerian_polynomials) for other definitions and several properties.
In particular, it is known the Eulerian polynomial has only negative and simple roots,
$$E_n(z) = \prod_{i=1}^n (1+q^{(n)}_i)$$
for different positive numbers $q^{(n)}_i$.

My question now is

> What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$.
This is,
$$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$
when they come in sorted order. That's *not* the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

$$
\begin{align*}
E_2(z) &= z^2 + 4z + 1
        = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\
E_3(z) &= z^3 + 11z^2 + 11z + 1
        = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1)
\end{align*}
$$

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$.
This can be used to show that
$$
\begin{align*}
  \sum_i \frac{1}{1+q^{(n)}_i}                     &= \frac{n}{2} \\
  \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12}
\end{align*}
$$

Doing this computation in the first example yields
$$
  \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1
$$
and
$$
  \frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3}
$$

> Is there any known immediate property of the roots that can be used to get these identities?