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# 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 .... 10 years ago

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

All the best.

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Sagar Singh

B.Tech, IIT Delhi

10 years ago

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.........