jmc

We begin by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to find the values of the two roots. From this, we get that $x=\frac{-5\pm \sqrt{25+12a}}{2a}$. Then, we can find $$x_1-x_2=\frac{-5+\sqrt{25+12a}}{2a}-\frac{-5-\sqrt{25+12a}}{2a}=\frac{\sqrt{25+12a}}{a}.$$Hence, $$\frac{\sqrt{12a+25}}{a}=\frac{\sqrt{61}}{3}.$$Squaring both sides, we get $$\frac{12a+25}{a^2}=\frac{61}{9},$$which simplifies to $61a^2-108a-225=0$. This equation factors as $(a-3)(61a+75)=0$. Since $a$ is positive, $a=\boxed{3}$.