jmc

If the quadratic expression on the left side has exactly one root in $x$, then it must be a perfect square. Dividing 9 from both sides, we have $x^2+\frac{n}{9}x+\frac{1}{9}=0$. In order for the left side to be a perfect square, it must factor to either ${\left(x+\frac{1}{3}\right)}^2=x^2+\frac{2}{3}x+\frac{1}{9}$ or ${\left(x-\frac{1}{3}\right)}^2=x^2-\frac{2}{3}x+\frac{1}{9}$ (since the leading coefficient and the constant term are already defined). Only the first case gives a positive value of $n$, which is $n=\frac{2}{3}\cdot 9=\boxed{6}$.