Question

If the sum of the roots of the equation a$x^2$ + bx + c = 0 is equal to the sum of the squares of their reciprocals, then $\frac{a}{c}$,$\frac{b}{a}$, $\frac{c}{b}$ are in :

• A. A.P.
• B. G.P.
• C. H.P.
• D. None

Answer 1

The correct option is C

H.P.

Suppose the roots are r and s. Then (x-r)(x-s) = 0, so
$x^2$ - (r+s) x + (rs) = 0
and hence r+s =$\frac{-b}{a}$ and rs = $\frac{c}{a}$

We are saying that
r + s = ($\frac{1}{r^2}$)+ ($\frac{1}{s^2}$)
<=> ${rs}^2$ (r+s) = $s^2$ +$r^2$
<=> ${rs}^2$ (r+s) =$(r+s{)}^2$- 2rs
<=> ($\frac{c}{a}{)}^2$ ($\frac{-b}{a}$)= ($\frac{-b}{a}{)}^2$- 2($\frac{c}{a}$)

Multiplying through by a³ we get
$-b{c}^2$= $a{b}^2$- 2$a^{2}c$
Divide through by abc to get
$\frac{-c}{a}$ =$\frac{b}{c}$ – 2$\frac{a}{b}$
<=> $\frac{a}{b}$ -$\frac{c}{a}$ = $\frac{b}{c}$ -$\frac{a}{b}$
which implies $\frac{c}{a}$ , $\frac{a}{b}$ and $\frac{b}{c}$ are in arithmetic progression.

Hence their reciprocals, $\frac{a}{c}$,$\frac{b}{c}$ and $\frac{c}{b}$ are in harmonic progression.