10.5. Determine all values of the real parameter $m$ for which the equation $m(m+2) \cdot x^{2}-(m-2) \cdot x\left(x^{2}+1\right)-2\left(x^{2}+1\right)^{2}=0$ has two distinct real solutions.
Solution. We observe that $x=0$ is not a solution to the equation. Dividing both sides of the equation by $x^{2}$, we obtain the equivalent equation

$$
2 \cdot\left(\frac{x^{2}+1}{x}\right)^{2}+(m-2) \cdot \frac{x^{2}+1}{x}-m(m+2)=0
$$

Let $\frac{x^{2}+1}{x}=t$, we get the equation $2 t^{2}+(m-2) \cdot t-m(m+2)=0$.

We will determine the set of values of $t$, that is, the set $E(f)$ of values of the function $f: \mathbb{R}^{*} \rightarrow \mathbb{R}$, $f(x)=\frac{x^{2}+1}{x}$.

$$
\frac{x^{2}+1}{x}=t \Leftrightarrow x^{2}-t x+1=0 . \quad \text { We calculate } \quad \Delta(t)=t^{2}-4
$$

$t \in E(f)$ if and only if $\Delta(t) \geq 0 \Leftrightarrow t \in(-\infty ;-2] \cup[2 ;+\infty)$, then $E(f)=(-\infty ;-2] \cup[2 ;+\infty)$.

(or $t=x+\frac{1}{x}$, from which for positive $x$ we have $t \geq 2$, and for negative $x$ we have $t \leq-2$.)

- For each value $t=2$ or $t=-2$ the equation $\frac{x^{2}+1}{x}=t$ has exactly one real solution. Equation (1) will have two real solutions if both numbers $t=2$ and $t=-2$ are simultaneously solutions of equation (2). This happens for $m=2$.
- For any value $t2$ the equation $\frac{x^{2}+1}{x}=t$ has exactly two real solutions.

For equation (2) we calculate $\Delta(m)=(m-2)^{2}+8 m(m+2)=9 m^{2}+12 m+4=(3 m+2)^{2} \geq 0$, from which equation (2) always has real solutions $\left[\begin{array}{l}t=-m, \\ t=\frac{m+2}{2}\end{array}\right.$.

1) Let $m=-\frac{2}{3}$. Then equation (2) has a single solution $t=\frac{2}{3} \notin(-\infty ;-2) \cup(2 ;+\infty)$. Therefore, in this case, equation (1) has no real solutions.
2) Let $m \in R \backslash\left\{-\frac{2}{3} ; 2\right\}$. Then equation (1) will have two distinct real solutions if equation (2) has only one solution in the interval $(-\infty ;-2) \cup(2 ;+\infty)$, and the other solution is in the interval $(-2 ; 2)$.

Thus $\left[\begin{array}{l}\left\{\begin{array}{l}-m \in(-\infty ;-2) \cup(2 ;+\infty) \\ \frac{m+2}{2} \in(-2 ; 2)\end{array}\right. \\ \left\{\begin{array}{l}\frac{m+2}{2} \in(-\infty ;-2) \cup(2 ;+\infty) \\ -m \in(-2 ; 2)\end{array} \Leftrightarrow\left[\begin{array}{l}m \in(-\infty ;-2) \cup(2 ;+\infty) \\ m \in(-6 ; 2)\end{array}\right.\right. \\ \left\{\begin{array}{l}m \in(-\infty ;-6) \cup(2 ;+\infty) \\ m \in(-2 ; 2)\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{c}m \in(-6 ;-2) \\ m \in \varnothing\end{array} \Leftrightarrow m \in(-6 ;-2)\right.\right.$.

Answer: $m \in(-6 ;-2) \cup\{2\}$.