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

Dear bhanu,

1 Given equation is ax² + bx + c = 0

2 Let α, β be the roots.

3 α + β = -ba and αβ = ca
 [Sum of the roots = -ba and Product of the roots = ca.]

4 α + β = 1 α2+ 1 β2
 [Given condition.]

5 α + β= α2 + β2 α2 β2
 [Simplify.]

6 α + β = (α + β)² - 2αβα² β² 
 [Use α² + β² = (α + β)² - 2α β.]

7 -ba =b²c² -2cac²a²
 [Substitute the values.]

8 -ba =b² - 2acc²
 [Multiply and divide by a² and simplify.]

9 -ba =b²c²-2acc²
 [Simplify.]

10 2acc²=b²c²+ba
 [Simplify.]

11 2a²c = ab² + bc²
 [Simplify.]

12 2ab=bc+ca
 [Divide both sides by abc.]

13 Therefore ab, bc, ca are in A.P

let A,B  be the roots of this equation then,

  A+B=-b/a  & AB=c/a          .................1

now it is given that sum of roots is equal to square of resiprocal of roots

so

         A+B=1/A² +1/B²

        A+B=A²+B² /(AB)² 

        A+B   = [(A+B)² -2AB]/(AB)²

now putting value of A+B and AB from eq 1

      -b/a =  [b²/a² -2c/a] /c²/a²

      ab² +bc² =2a²c                      ........................2

now let a/c,b/a,c/b are in HP then

           a/b = [b/c + c/a]/2

        ab² + bc² =2a²c

this result is similar to eq 2 so these terms are in HP.........