10.8. Determine all values of the real parameter $a$, for which the equation $2 \cdot|2 \cdot| x\left|-a^{2}\right|=x-a$ has exactly three real solutions.
Solution. If we consider $a=0$, then we obtain the equation $4 \cdot|x|=x \Leftrightarrow x=0$.

For $a=0$ the equation has only one real solution, so $a \neq 0$.

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x-a$. For any fixed value of the parameter $a$, the graph of the function $f$ is a line with a slope equal to 1, which is parallel to the support line of the bisectors of the first and third quadrants. The graph of the function $f$ intersects the x-axis at the point $(a ; 0)$.

Consider the function $g: \mathbb{R} \rightarrow \mathbb{R}, g(x)=2 \cdot|2 \cdot| x\left|-a^{2}\right|$. We observe that the function $g$ is even. We will draw the graph of the function $g$ for $x \geq 0$, and then, through a symmetry with respect to the $O y$ axis, we will draw the graph of the function $g$.

For $x \geq 0$, we obtain $g(x)=2 \cdot\left|2 x-a^{2}\right|=\left\{\begin{array}{l}-4 x+2 a^{2} \text {, if } 0 \leq x<\frac{a^{2}}{2} \\ 4 x-2 a^{2} \text {, if } x \geq \frac{a^{2}}{2}\end{array}\right.$.

Then we can write:

$g: \mathbb{R} \rightarrow \mathbb{R}, g(x)=\left\{\begin{array}{l}-4 x-2 a^{2}, \text { if } \quad x<-\frac{a^{2}}{2} \\ 4 x+2 a^{2}, \text { if }-\frac{a^{2}}{2} \leq x<0 \\ -4 x+2 a^{2}, \text { if } 0 \leq x<\frac{a^{2}}{2} \\ 4 x-2 a^{2}, \text { if } x \geq \frac{a^{2}}{2}\end{array}\right.$.

We observe that the graph of the function $g$ is the union of two rays and two segments. The support lines of these have slopes of -4 or 4.

The initial equation will have three distinct real solutions if and only if the graphs of the functions $f$ and $g$ will have three distinct common points. This will happen only if the graph of the function $f$ will pass through the point $\left(-\frac{a^{2}}{2} ; 0\right)$ or through the point $(0 ; g(0))$, i.e., through the point $\left(0 ; 2 a^{2}\right)$.

Therefore, we solve the following:

![](https://cdn.mathpix.com/cropped/2024_06_06_f4f7cfcf3dfd8b0307cag-3.jpg?height=1122&width=899&top_left_y=972&top_left_x=1024)
$\left[\begin{array}{l}f\left(-\frac{a^{2}}{2}\right)=0 \\ f(0)=2 a^{2}\end{array} \Leftrightarrow\left[\begin{array}{l}-\frac{a^{2}}{2}-a=0 \\ -a=2 a^{2}\end{array} \stackrel{a \neq 0}{\Leftrightarrow}\left[\begin{array}{l}a=-2 \\ a=-\frac{1}{2}\end{array}\right.\right.\right.$.

Answer: $a \in\left\{-2 ;-\frac{1}{2}\right\}$.