Mathematics - Quadratic Equation Question with Solution | TestHub

$a{x}^2+bx+c=0,\alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$ As for given condition, $\alpha +\beta =\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ $\alpha +\beta =-\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}-\frac{b}{a}=\frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}}$ On simplification $2a^{2}c=a{b}^2+b{c}^2$ $\Rightarrow \frac{2\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{b}}{\mathrm{c}}\therefore \frac{\mathrm{a}}{\mathrm{b}},\frac{\mathrm{b}}{\mathrm{a}},\&\frac{\mathrm{c}}{\mathrm{b}}$ are in H.P.