Question

what is the sum of the reciprocals of the solutions of x^3-3x^2-13x+15=0?

Answer 1

@mitbrij see this - http://mathcountsnotes.blogspot.in/2010/12/sum-and-product-super-stretch-on-page.html

Answer 2

If \(r_1, r_2, r_3\) are the roots, then
\[\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{r_2r_3+r_1r_3+r_1r_2}{r_1r_2r_3}\]
By Viete's formulas, \(r_2r_3+r_1r_3+r_1r_2=-13\) and \(r_1r_2r_3=15\).

Answer 3

Oops. \(r_1r_2r_3=-15\), not 15.

Answer 4

Solve for the roots. since the factors of 15 are 3, 5 or 15, 1. The most likely candidate is the 3,5 using synthetic division or long division you will find that (x+3) is one of the factors, dividing x^3-3x^2-13x+5 by (x+3) gives you x^2-6x+5 which factors
to (x-5)(x-1) we have now the three factors (x+3), (x-5)(x-1) the roots then are:
x=-3, x=5, x=1 giving you reciprocals of -1/3, 1, 1/5 summing them:
-1/3 +1/5 + 1= -5/15 + 3/15 + 15/15 =13/15