jmc

Let the three roots of the equation be $a,$ $a,$ and $b.$ Then by Vieta's formulas, $$\begin{array}{rl}a+a+b& =-\frac{8}{2}=-4,\\ ab+ab+a^2& =\frac{120}{2}=-60.\end{array}$$These equations simplify to $2a+b=-4$ and $2ab+a^2=-60.$ From the first equation, we get $b=-4-2a,$ and substituting into the second equation gives $$2a(-4-2a)+a^2=-60,$$or $$3a^2+8a-60=0.$$This factors as $$(a+6)(3a-10)=0,$$so either $a=-6$ or $a=\frac{10}{3}.$ If $a=-6$, then $b=-4-2a=8,$ so by Vieta, $k=-2a^{2}b=-576,$ which is not positive. If $a=\frac{10}{3},$ then $b=-4-2a=-\frac{32}{3},$ so by Vieta, $k=-2a^{2}b=\boxed{\frac{6400}{27}},$ which is the answer.