Find the sum of the reciprocals of the roots of $x^2-13x+4=0$.
I know that the sum of the roots of a quadratic equation is equal to the opposite of the coefficient of the linear term divided by the coefficient of the quadratic term, by Vieta's formulas. So, if the roots are $r_1$ and $r_2$, then $r_1+r_2=13/1=13$. I also know that the product of the roots of a quadratic equation is equal to the constant term divided by the coefficient of the quadratic term, by Vieta's formulas. So, $r_1r_2=4/1=4$. To find the sum of the reciprocals of the roots, I can write $\frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1+r_2}{r_1r_2}$. Substituting the values of $r_1+r_2$ and $r_1r_2$, I get $\frac{1}{r_1}+\frac{1}{r_2}=\frac{13}{4}$. 