For two real values of $n$, the equation $9x^2+nx+36=0$ has exactly one solution in $x$. What is the positive value of $n$?
To have exactly one solution, the quadratic equation must have a zero discriminant, that is, $b^2-4ac=0$. In this case, $a=9, b=n, c=36$, so I need to find $n$ such that $n^2-4\cdot 9\cdot 36=0$. Expanding and simplifying, I get $n^2-1296=0$. To solve for $n$, I can take the square root of both sides, remembering the plus or minus sign. I get $n=\pm\sqrt{1296}$. Since I only want the positive value of $n$, I choose the plus sign and simplify. I get $n=\sqrt{1296}=36$. 