jmc

By considering the expression $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for the solutions of $a{x}^2+bx+c=0$, we find that the solutions are rational if and only if the discriminant $b^2-4ac$ has a rational square root. Therefore, the solutions of $k{x}^2+20x+k=0$ are rational if and only if $400-4(k)(k)$ is a perfect square. (Recall that if $n$ is an integer which is not a perfect square, then $\sqrt{n}$ is irrational). By writing the discriminant as $4(100-k^2)$, we see that we only need to check the integers $1\le k\le 10$. Of these, $\boxed{6,8\text{, and }10}$ work.