Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$?
Let us write \begin{align}H_n(x)&=A_n(x-x_1)\ldots(x-x_n)=\\ &=A_n\left(x^n-e_1(x_1,\ldots,x_n)x^{n-1}+e_2(x_1,\ldots,x_n)x^{n-2}+\mathrm{poly}_{n-3}(x)\right),\tag{1} \end{align} where $e_k(x_1,\ldots,x_n)$ denote elementary symmetric polynomials: \begin{align} & e_1(x_1,\ldots,x_n)=\sum_{k=1}^nx_k,\\ & e_2(x_1,\ldots,x_n)=\sum_{1\leq i<j\leq n}^n x_ix_j. \end{align}
We want to find $$\sum_{k=1}^{n}x_k^2=e_1^2(x_1,\ldots,x_n)-2e_2(x_1,\ldots,x_n),\tag{2}$$ and therefore it will suffice to know the coefficients of $x^n$, $x^{n-1}$ and $x^{n-2}$ in $H_n(x)$. But they can be determined from the series representations of Hermite polynomials: $$H_n(x)=2^n\left(x^n-\frac{n(n-2)}{4}x^{n-2}+\mathrm{poly}_{n-4}(x)\right).\tag{3}$$
Together with (1) and (2), this gives the result: $$\sum_{k=1}^{n}x_k^2=\frac{n(n-1)}{2}.$$