Question

If the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is equal to the sum of 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. A

Answer 1

The correct option is C H.P.

$a{x}^2+bx+c=0$

Let $\alpha ,\beta$ be the roots of the equation

$\Rightarrow \alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$ ………..$\left(1\right)$

$\Rightarrow \alpha +\beta =\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ [Given]

$\Rightarrow \alpha +\beta =\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$

$\Rightarrow \alpha +\beta =\frac{(\alpha +\beta {)}^2-2\alpha \beta }{{\alpha }^2{\beta }^2}$ ………….$\left(2\right)$ substituting the values of equation $\left(1\right)$ in equation $\left(2\right)$

$\Rightarrow \frac{-b}{a}=\frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}}$

$\Rightarrow \frac{-b}{a}=\frac{b^2-2ac}{c^2}$

$\Rightarrow -b{c}^2=a{b}^2-2a^{2}c$

$\Rightarrow 2a^{2}c=a{b}^2+b{c}^2$

Then $a{b}^2,a^{2}c,b{c}^2$ are in A.P

Dividing by abc, we get:-

$\frac{b}{c},\frac{a}{b},\frac{c}{a}$ are in AP

$\Rightarrow \frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in AP

$\Rightarrow \frac{a}{c},\frac{b}{a},\frac{c}{b}$ are in HP.