Find a quadratic polynomial whose zeroes …

$\text { Let } \alpha \text { and } \beta \text { are the zeroes of the polynomial } f(x)=a x^2+b x+c \text {. }$

$\therefore \quad(\alpha+\beta)=\frac{-\mathbf{b}}{\mathbf{a}} and \alpha \beta=\frac{\mathbf{c}}{\mathbf{a}} According to the question, \frac{1}{\alpha} and \frac{1}{\beta} are the zeroes of the required quadratic polynomial$

Sum of zeroes of required polynomial

$\begin{aligned} \mathbf{S}^{\prime} & =\frac{1}{\alpha}+\frac{1}{\beta} \\ & =\frac{\alpha+\beta}{\alpha \beta} \\ & =\frac{-\mathbf{b}}{\mathbf{c}} \end{aligned} [From equation (i) and (ii)] and product of zeroes of required polynomial =\frac{1}{\alpha} \times \frac{1}{\beta}. P^{\prime}=\frac{1}{\alpha \beta} =\frac{a}{c} [From equation (ii)]$

Equation of the required polynomial.

$=\mathbf{k}\left(\mathbf{x}^2-\mathbf{S}^{\prime} \mathbf{x}+\mathbf{p}^{\prime}\right) where $k$ is any non-zero constant =k\left(x^2-\left(\frac{-b}{c}\right) x+\frac{a}{c}\right) [From equation (iii) and (iv)] =k\left(x^2+\frac{b}{c} x+\frac{a}{c}\right)$