Given $x\neq0$, find the positive value of $b$ such that the equation $\frac 3x+\frac x3=b$ will have exactly one solution.
I notice that the equation involves fractions, so I want to clear them by multiplying both sides by the lowest common denominator, which is $3x$. This gives me $9+x^2=3bx$, which I can rearrange to get a quadratic equation in terms of $x$: $x^2-3bx+9=0$. Now, I know that a quadratic equation has exactly one solution when its discriminant, which is the part under the square root in the quadratic formula, is equal to zero. The discriminant is $(-3b)^2-4(1)(9)$, so I want to solve $(-3b)^2-4(1)(9)=0$ for $b$. Expanding and simplifying, I get $9b^2-36=0$, which I can divide by $9$ to get $b^2-4=0$. Taking the square root of both sides, I get $b=\pm2$. However, the problem asks for the positive value of $b$, so I can discard the negative value and conclude that $b=2$. 