Problem 8. For what values of the parameter $a$ does the equation

$$
\log _{3}\left(2 x^{2}-x+2 a-4 a^{2}\right)+\log _{1 / 3}\left(x^{2}+a x-2 a^{2}\right)=0
$$

have two distinct roots, the sum of whose squares is less than 1?
Answer: $\left(0 ; \frac{1}{3}\right) \cup\left(\frac{1}{3} ; \frac{2}{5}\right)$.

Solution. Rewrite the original equation as

$$
\log _{3}\left(2 x^{2}-x+2 a-4 a^{2}\right)=\log _{3}\left(x^{2}+a x-2 a^{2}\right)
$$

Notice that this equation is equivalent to the system

$$
\left\{\begin{array}{l}
2 x^{2}-x+2 a-4 a^{2}=x^{2}+a x-2 a^{2} \\
x^{2}+a x-2 a^{2}>0
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x^{2}-(a+1) x-2 a(a-1)=0 \\
(x-a)(x+2 a)>0
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-16.jpg?height=468&width=437&top_left_y=1542&top_left_x=1529)

By the converse of Vieta's theorem, the roots of equation (1) are $x_{1}=2 a$ and $x_{2}=-a+1$. Since the equation must have two distinct roots, we get the following conditions:

$$
\left\{\begin{array}{l}
(2 a-a)(2 a+2 a)>0 \\
(-a+1-a)(-a+1+2 a)>0 \\
2 a \neq-a+1
\end{array}\right.
$$

that is, $a \neq 0, a \neq \frac{1}{3}, a \in\left(-1 ; \frac{1}{2}\right)$.

The sum of the squares of $x_{1}$ and $x_{2}$ must be less than 1, that is, $4 a^{2}+(a-1)^{2}<1,5 a^{2}-2 a<0$, from which $a \in\left(0 ; \frac{2}{5}\right)$. Intersecting all four obtained conditions, we get the answer.