Question

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

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

Answer 1

Answer: (A)

A.P

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

$\Rightarrow \alpha +\beta =\frac{-b}{a}$

$\Rightarrow \alpha \beta =\frac{c}{a}$

Given that

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

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

Substituing these values we get

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

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

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

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

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

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

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

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

Hence $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in $A.P$