jmc

By the quadratic formula, the roots of the equation are $$\begin{array}{rl}\frac{-b\pm \sqrt{b^2-4ac}}{2a}& =\frac{-k\pm \sqrt{k^2-4(5)(1)}}{2(1)}\\ & =\frac{-k\pm \sqrt{k^2-20}}{2}.\end{array}$$ We want the difference of the roots, so we take the larger minus the smaller: $$\begin{array}{rl}\left(\frac{-k+\sqrt{k^2-20}}{2}\right)-\left(\frac{-k-\sqrt{k^2-20}}{2}\right)& =\frac{2\sqrt{k^2-20}}{2}\\ & =\sqrt{k^2-20}.\end{array}$$ We are given that this difference is equal to $\sqrt{61}$, so we have $$\begin{array}{rl}\sqrt{k^2-20}& =\sqrt{61}\Rightarrow \\ k^2-20& =61\Rightarrow \\ k^2& =81\Rightarrow \\ k& =\pm 9.\end{array}$$ Thus the greatest possible value of $k$ is $\boxed{9}$.