2.40 Form a second-degree equation, one of whose roots is equal to the sum, and the other is equal to the product of the roots of the equation $a x^{2}+b x+c=0$.
2.40 If the roots of the desired equation are denoted by $\bar{x}_{1}$ and $\bar{x}_{2}$, and those of the given equation by $x_{1}$ and $x_{2}$, then from the condition it follows that $\bar{x}_{1}=x_{1}+x_{2} ; \bar{x}_{2}=x_{1} x_{2}$. Using Vieta's theorem, we get $\bar{x}_{1}=-\frac{b}{a} ; \bar{x}_{2}=\frac{c}{a}$. Hence, $\bar{x}_{1}+\overline{x_{2}}=\frac{c-b}{a} ; \bar{x}_{1} \bar{x}_{2}=-\frac{b c}{a^{2}}$. Therefore, the desired equation will take the form

$$
x^{2}+\frac{b-c}{a} x-\frac{b c}{a^{2}}=0, \text { or } a^{2} x^{2}+(a b-a c) x-b c=0
$$