Explanation

(d)
Let roots of the equation $ax^{2} + bx + c = 0$ are $\alpha\$and$\ \beta$
$\therefore\ \ \alpha + \beta = - \frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$
Now, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$
$= \frac{- \frac{b}{a}}{\frac{c}{a}} = \frac{- b}{c}$
And $\frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\frac{c}{a}} = \frac{a}{c}$
$\therefore$ Required equation is
$x^{2} - \left( - \frac{b}{c} \right)x + \frac{a}{c} = 0$
$\Rightarrow \ \ cx^{2} + bx + a = 0$
Alternate To find the equation of reciprocal rots, interchange the coefficients of $x^{2}$ and constant term in the given equation then required equation is $cx^{2} + bx + a = 0$