4. Find all values of the parameter $a$ for which the equation

$$
\left(\sqrt{6 x-x^{2}-4}+a-2\right)((a-2) x-3 a+4)=0
$$

has exactly two distinct real roots.

(P. Alishev)
Answer: $a \in\{2-\sqrt{5}\} \cup\{0\} \cup\{1\} \cup\left(2-\frac{2}{\sqrt{5}} ; 2\right]$

Solution: We will solve the problem graphically. Replace $a$ with $y$ and draw the set of solutions of the equation in the $Oxy$ plane.

![](https://cdn.mathpix.com/cropped/2024_05_06_9da091ab250974b7674ag-4.jpg?height=794&width=1513&top_left_y=360&top_left_x=334)

The original equation is equivalent to the following system:

$$
\left[\begin{array} { l } 
{ y = 2 - \sqrt { 6 x - x ^ { 2 } - 4 } , } \\
{ ( y - 2 ) x - 3 y + 4 = 0 ; }
\end{array} \Longrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
(x-3)^{2}+(y-2)^{2}=5 \\
y \leqslant 2
\end{array}\right. \\
(x-3)(y-2)=2
\end{array}\right.\right.
$$

The first equation defines a semicircle with center at point $(3 ; 2)$, and the second equation defines a hyperbola with two asymptotes $x=3$ and $y=2$ (see the figure). Also, we should not forget about the domain: $6 x-x^{2}-4 \geqslant 0 \Longrightarrow x \in[3-\sqrt{5} ; 3+\sqrt{5}]$.

In the end, we get that the original equation has exactly two solutions if and only if the line $y=a$ intersects the semicircle and the hyperbola exactly at two points in the strip $3-\sqrt{5} \leqslant x \leqslant 3+\sqrt{5}$. From the graph, we can see that there are three possible cases: the line is tangent to the semicircle (i.e., passes through point $D$); the line passes through one of the points of intersection of the semicircle with the hyperbola (i.e., through point $A$ or $B$); the line lies strictly above the line passing through point $C$ and not strictly below the line $y=2$. Let's consider all these three cases:

I) Find the $y$-coordinate of point $D$:

$$
y=2-\sqrt{6 x-x^{2}-4}=2-\sqrt{6 \cdot 3-3^{2}-4}=2-\sqrt{5} .
$$

II) Find the $y$-coordinates of points $A$ and $B$:

$$
\left\{\begin{array} { l } 
{ ( x - 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 5 , } \\
{ ( x - 3 ) ( y - 2 ) = 2 , } \\
{ y \leqslant 2 ; }
\end{array} \Longrightarrow \left\{\begin{array}{l}
(y-2)^{4}-5(y-2)^{2}+4=0 \\
y \leqslant 2 ;
\end{array}\right.\right.
$$

Factorize the first equation or solve it as a biquadratic equation:

$$
\left\{\begin{array}{l}
(y-4) y(y-3)(y-1)=0, \\
y \leqslant 2 ;
\end{array} \quad \Longrightarrow y=0 \text { or } y=1 .\right.
$$

III) Find the $y$-coordinate of point $C$:

$$
\begin{array}{r}
(x-3)(y-2)=2 \Longrightarrow((3-\sqrt{5})-3)(y-2)=2 \Longrightarrow \\
\Longrightarrow y=2-\frac{2}{\sqrt{5}}
\end{array}
$$

Thus, we get the following values of the parameter:

$$
y \in\{2-\sqrt{5}\} \cup\{0\} \cup\{1\} \cup\left(2-\frac{2}{\sqrt{5}} ; 2\right] .
$$