Task 2. Find the set of values of the parameter $a$ for which the discriminant of the equation $a x^{2}+2 x+1=0$ is 9 times the square of the difference between its two distinct roots.
Solution. $D=4-4 a$.

$$
\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=\left(\frac{2}{a}\right)^{2}-4 \cdot \frac{1}{a}=\frac{4-4 a}{a^{2}}=\frac{D}{a^{2}}
$$

We obtain the equation: $\frac{D}{a^{2}} \cdot 9=D$. The condition $D>0$ is satisfied only by the root $a=-3$.

Answer: $a \in\{-3\}$.