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 $k{x}^2+10x+k=0$ are rational if and only if $100-4(k)(k)$ 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 $4(25-k^2)$, we see that we only need to check the integers $1\le k\le 5$. Of these, 3, 4, and 5 work, for a total of $\boxed{3}$ integer values of $k$.