The sume of the reciprocals of the roots of x^3+ax^2+bx+c=0 is:?

Question

(a/b), (a/c), (-a/c), (-b/c), or (1/(a+c))

Answer 1

I am going with -b/c

Theorem: if (x-r1)(x-r2)(x-r3) = x^3+ax^2+bx+c then a= -(r1+r2+r3) while b=r1*r2+ r1*r3+r2*r3 and c= -r1*r2*r3 so then 1/r1+1/r2+1/r3 = (r2*r3+r1*r3 +r1*r2)/(r1*r2*r3) common denominator! so this equals b/(-c)

If you aren't too interested in the theorem stuff, you can pick roots and see which one checks out. So for instance distribute out (x-2)(x-3)(x-5) to find a,b,c then see which of the formulas is equal to 1/2+1/3+1/5 ......