Roots of polynomial $\sum_{k = 0}^{n} \sqrt{\binom{n}{k}} x^k = 0$ I'm interested in roots of the polynomial $$p(x) = \sum_{k = 0}^{n} \sqrt{\binom{n}{k}}  x^k = 0$$
I've looked around but can't find anything in the literature or forums. Since coefficients are all real, roots come in complex conjugate pairs. Having looks at some solutions for specific values of $n$ I also conjecture that for certain values of $n$ all roots lie on $|x| = 1$, but have no idea how to approach this (eg for $n=11$ all but 2 roots lie on the unit circle). Any ideas would be appreciated.
 A: This isn't an answer but some remarks that can help for making conjectures.
Let $$p_n(x) := \sum_{k = 0}^{n} \sqrt{\binom{n}{k}}  x^k $$
The fact that the non-real roots of many polynomials are close to the unit circle has been remarked and studied many times for example here, but here they are on the unit circle but some "outsiders" (or "insiders" :)) (see graphical representation below).
But $p_n$ has the property of being a self-reciprocal (or palindromic) polynomial:
$$x^{n}p_n(\tfrac1x)=p_n(x) \tag{1}$$
The fact that the roots of such polynomials can be all on the unit circle has been studied, for example here, but we are here in a case where a  few of them are not on it...

Fig. 1: The roots are the little red points with the index of their resp. polynomials. You will find for example 5 roots labelled "5".
Some elementary remarks:

*

*The roots come by conjugate pairs because the coefficients of $p_n$ are real.


*Root $-1$ is common to all $p_n$ with $n$ odd, precisely because $p_n$ is a self reciprocal polynomial : indeed take $x=-1$ in (1).
