9. Determine all values of $a$ for which the equation

$$
4 x^{2}-8|x|+(2 a+|x|+x)^{2}=4
$$

has exactly two distinct roots. Specify these roots for each of the found values of $a$.

(12 points)
# Solution:

I. $x \geq 0, x^{2}-2 x+(a+x)^{2}=1 ; 2 x^{2}+2(a-1) x+a^{2}-1=0$ ；

$D / 4=1-2 a+a^{2}-2 a^{2}+2=3-2 a-a^{2}$.

The equation has two distinct non-negative roots $x_{1,2}=\left(1-a \pm \sqrt{3-2 a-a^{2}}\right) / 2$, if $\left\{\begin{array}{c}3-2 a-a^{2}>0, \\ a-10\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\left[\begin{array}{l}a=-3, \\ a=1, \\ a \leq 1,\end{array}\right.\right. \\ -10, \\ a^{2}-1>0, \\ -1<0\end{array} \Leftrightarrow\left[\begin{array}{c}-\sqrt{2}<a<-1, \\ 1<a<\sqrt{2} .\end{array}\right.\right.$

The equation has exactly one negative root $x=-1-\sqrt{2-a^{2}}$, if $\left[\begin{array}{l}\left\{\begin{array}{c}2-a^{2}=0, \\ -1<0,\end{array}\right. \\ a^{2}-1<0, \\ \left\{\begin{array}{c}a^{2}-1=0, \\ -1<0\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{c}a=-\sqrt{2}, \\ a=\sqrt{2}, \\ -1 \leq a \leq 1 .\end{array}\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_18575ac96ab43cb21cd9g-07.jpg?height=242&width=1126&top_left_y=1895&top_left_x=545)

Answer: $a \in(-3 ;-\sqrt{2}), \quad x_{1,2}=\left(1-a \pm \sqrt{3-2 a-a^{2}}\right) / 2$;

$$
\begin{aligned}
& a \in(-1 ; 1], x_{1}=\left(1-a+\sqrt{3-2 a-a^{2}}\right) / 2, x_{2}=-1-\sqrt{2-a^{2}} \\
& a \in(1 ; \sqrt{2}), x_{1,2}=-1 \pm \sqrt{2-a^{2}}
\end{aligned}
$$