TITLE: Find the sum of the inverse roots of $x^3-7x^2+4x-1$
QUESTION [5 upvotes]: Calculate the sum of the inverse roots of the equation $x^3-7x^2+4x-1=0$.
My development was:
sum = $-7$
product = $1$
Thus, I believe that to find the inverse roots one only has to share the sum with the product, i.e. $-\frac{7}{1} = 7$, however, the answer given in the question is $4$.
what do you think? do you have any formula?

REPLY [8 votes]: Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$\frac1{x_1}+\frac1{x_2}+\frac1{x_3}=\frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=\frac{-c/a}{d/a}=-\frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.