If the sum of roots of equation ax^2+bx+c=0 is equal to sum of the square of their reciprocals then a/c,b/a,c/b are in?

• let p,q be the roots then p+q=-(b/a)=(1/p^2+1/q^2)so(p^2+q^2)/(pq)^2=-(b/a)then{(p+q)^2-2*p*q}/(pq)^2=-(b/a)now pq=c/a; so,putting values we get{(-b/a)^2-2*(c/a)}/(c/a)^2)=-(b/a) now dividing both side by abc we finally get (b/c)+(c/a)=2*(a/b)so b/c,a/b,c/a are in a.p...so a/c,b/a,c/b are in h.p
  
• 11 years agoHelpfull: Yes(24) No(0)