2.39 Form a quadratic equation with roots $\frac{1}{x_{1}}$ and $\frac{1}{x_{2}}$, where $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$.
### 2.39 We have

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}} ; \frac{1}{x_{1}} \cdot \frac{1}{x_{2}}=\frac{1}{x_{1} x_{2}}
$$

By Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a} ; x_{1} x_{2}=\frac{c}{a}$. Therefore, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c} ; \frac{1}{x_{1}} \cdot \frac{1}{x_{2}}=\frac{a}{c}$.

Thus, $x^{2}+\frac{b}{c} x+\frac{a}{c}=0$, or $c x^{2}+b x+a=0$.