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

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

have two distinct roots, the sum of whose squares is greater than 4?
Answer: $(-\infty ;-1) \cup\left(\frac{3}{5} ; 1\right)$.

Solution. Rewrite the original equation as

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

Notice that this equation is equivalent to the system

$$
\left\{\begin{array}{l}
2 x^{2}+(2 a+1) x-2 a=x^{2}+3 a x+2 a^{2} \\
x^{2}+3 a x+2 a^{2}>0
\end{array}\right.
$$

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-04.jpg?height=466&width=423&top_left_y=1543&top_left_x=1542)

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 obtain 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, $a \neq 0, a \neq-\frac{1}{3}, 4,5 a^{2}+2 a-3>0$, from which $a \in(-\infty ;-1) \cup\left(\frac{3}{5} ; \infty\right)$. Intersecting all four obtained conditions, we get the answer.