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

$$
\log _{2}\left(2 x^{2}-x-2 a-4 a^{2}\right)+3 \log _{1 / 8}\left(x^{2}-a x-2 a^{2}\right)=0
$$

have two distinct roots, the sum of whose squares lies in the interval $(4 ; 8)$?
Answer: $\left(\frac{3}{5} ; 1\right)$.

Solution. Rewrite the original equation as

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

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-22.jpg?height=474&width=434&top_left_y=1536&top_left_x=1531)

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:

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

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

The sum of the squares of $x_{1}$ and $x_{2}$ must belong to the interval $(4 ; 8)$, that is

$$
\left\{\begin{array}{l}
5 a^{2}+2 a-7>0 \\
5 a^{2}+2 a-7<0
\end{array}\right.
$$

from which $a \in(-\infty ;-1) \cup\left(\frac{3}{5} ;+\infty\right), a \in\left(-\frac{7}{5} ; 1\right)$. Intersecting all five obtained conditions, we get the answer.