4. Find all values of the parameter $\alpha$, for each of which the system

$$
\left\{\begin{array}{c}
\sqrt{-5 y+4+4 \alpha}-\sqrt{\alpha-x}=0 \\
y-\sqrt{x}=0
\end{array}\right.
$$

has at least one solution with respect to the variables $y$ and $x$.
Solution. Taking into account the domain of definition (ODZ) for this problem, the original system transforms into:

$$
\left\{\begin{array}{c}
-5 y+4+4 \alpha \geq 0 \\
\alpha-x \geq 0 \\
y \geq 0 \\
x \geq 0 \\
-5 y+4+4 \alpha=\alpha-x \\
y=\sqrt{x}
\end{array}\right.
$$

Let's explicitly express the constraints on the variables $y$ and $x$. Also, in the first equality, express $x$ in terms of $y$. We get:

$$
\left\{\begin{array}{c}
0 \leq y \leq \frac{4(1+\alpha)}{5} \\
0 \leq x \leq a \\
x=5 y-4-3 \alpha \\
y=\sqrt{x}
\end{array}\right.
$$

Substitute $5 y-4-3 \alpha$ for $x$ in the last equality and square both sides. We get:

$$
y^{2}=5 y-4-3 \alpha
$$

This quadratic equation has at least one solution if its discriminant is non-negative. We have the inequality $D=(-5)^{2}-4(4+3 \alpha)>0$. This means the quadratic equation has a solution only for $\alpha \leq \frac{3}{4}$. Then the roots of the equation are:

$$
y_{1,2}=\frac{1}{2}(5 \pm \sqrt{9-12 \alpha})
$$

Let's check if these roots are less than or equal to $\frac{4(1+\alpha)}{5}$. First, check for $y_{1}$:

$$
\begin{gathered}
\frac{1}{2}(5+\sqrt{9-12 \alpha}) \leq \frac{4(1+\alpha)}{5} \\
(25+5 \sqrt{9-12 \alpha}) \leq 8 \alpha+8 \\
5 \sqrt{9-12 \alpha} \leq 8 \alpha-17
\end{gathered}
$$

In this inequality, the left side is greater than zero and defined since $\alpha \leq \frac{3}{4}$. Therefore, for this inequality to hold, it is necessary that $8 \alpha-17 \geq 0 \Rightarrow \alpha \geq \frac{17}{8}$. This leads to a contradiction. Now, check the inequality for $y_{2}$:

$$
\begin{gathered}
\frac{1}{2}(5-\sqrt{9-12 \alpha}) \leq \frac{4(1+\alpha)}{5} \\
(25-5 \sqrt{9-12 \alpha}) \leq 8 \alpha+8 \\
-5 \sqrt{9-12 \alpha} \leq 8 \alpha-17 \\
5 \sqrt{9-12 \alpha} \geq 17-8 \alpha \\
25(9-12 \alpha) \geq(8 \alpha-17)^{2} \\
64 \alpha^{2}-272 \alpha+289 \leq 225-300 \alpha \\
64 \alpha^{2}+28 \alpha+64 \leq 0
\end{gathered}
$$

| Criteria for evaluating the completion of the task | Points |
| :---: | :---: |
| A correct and justified sequence of all steps of the solution. All transformations and calculations are performed correctly. The correct answer is obtained. | 10 |
| A correct sequence of all steps of the solution. Gaps in the justification of the solution (for example, when comparing the obtained roots with the domain of definition) or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. | 6 |
| The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the domain of definition is correctly found and the quadratic equation $y^{2}=5 y-4-3 \alpha$ is obtained); - the other part is either not performed or performed incorrectly, | 3 |

The quadratic equation $64 \alpha^{2}+28 \alpha+64=0$ has a negative discriminant. Since the branches of the parabola are directed upwards, there are no $\alpha$ such that $64 \alpha^{2}+28 \alpha+64 \leq 0$. This leads to a contradiction. This means that there are no $\alpha$ for which there is at least one root of the equation $y^{2}=5 y-4-3 \alpha$ and the conditions of the domain of definition are satisfied.

Answer: No solutions.
It is possible that even with logical errors (for example, the roots of the quadratic equation $y^{2}=5 y-4-3 \alpha$ are found incorrectly, the obtained roots are compared incorrectly with the domain of definition). In this case, the solution may be incomplete.

All cases of the solution that do not meet the above criteria for scoring 3 and 6 points.