problem: If the quadratic $x^2+6mx+m$ has exactly one real root, find the positive value of $m$.
solution: Consider the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Since the quadratic has exactly one root, then its discriminant must be 0. Thus, this gives us  \begin{align*} 0&=b^2-4ac
  \\\Rightarrow\qquad0&=(6m)^2-4m
  \\\Rightarrow\qquad0&=36m^2-4m
  \\\Rightarrow\qquad0&=4m(9m-1).
  \end{align*}This gives us the two possible values of $m$: 0 and $\frac{1}{9}$. Since the question only asks for the positive value, our final answer is $\boxed{\frac19}$.