problem: The two solutions of the equation $x^2+bx+48=0$ are in the ratio of 3 to 1 for some values of $b$. What is the largest possible value of $b$?
solution: For this problem, we make use of the correspondence between sums/products of roots and coefficients of a polynomial.
  
  Denote the two roots of the equation $\alpha$ and $\beta$. We know that $\alpha\beta = 48$, and $\alpha/\beta = 3 \implies \alpha = 3\beta$.
  
  So $ b = -\alpha - \beta = -4\beta$. To maximize $b$, we want to make $\beta$ negative and as large as possible. Given the relationship that $\alpha = 3\beta$ and that $\alpha*\beta = 48$, we see that $\beta = 4$ or $-4$. Clearly $-4$ maximizes $b$, and $b = \boxed{16}$.