Question

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

• A. arithmetic progression
• B. geometric progression
• C. harmonic progression
• D. arithmetico-geometric progression

Answer 1

Answer: (C)

Given equation is
$a{x}^2+bx+c=0$
Let $\alpha ,\beta$ be the roots of this equation.
then, $\alpha +\beta =-\frac{b}{a}$ and $a\beta =\frac{c}{a}$
Also, $\alpha +\beta =\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{a^2+{\beta }^2}{{\alpha }^2{\beta }^2}$
$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{(\alpha \beta {)}^2}$
$\Rightarrow \alpha +\beta ={\left(\frac{\alpha +\beta }{\alpha \beta }\right)}^2-\frac{2}{\alpha \beta }$
$\Rightarrow \left(-\frac{b}{a}\right)={\left(\frac{-b/a}{c/a}\right)}^2-\frac{2}{c/a}$
$\Rightarrow -\frac{b}{a}={\left(\frac{b}{c}\right)}^2-\frac{2a}{c}$
$\Rightarrow \frac{2a}{c}={\left(\frac{b}{c}\right)}^2+\frac{b}{a}$
$\Rightarrow \frac{2a}{c}=\frac{b}{c}\left[\frac{b}{c}+\frac{c}{a}\right]$
$\Rightarrow \frac{2a}{b}=\frac{b}{c}+\frac{c}{a}$
$\Rightarrow \frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in A.P.
$\Rightarrow \frac{a}{c},\frac{b}{a},\frac{c}{b}$ are in H.P.