2. Find all real numbers $x$ for which the value of the expression

$$
\sqrt{1-x^{2}}+\sqrt{5 x-x^{2}}
$$

is an integer.
IV/2. Estimate the value of the expression. Clearly, $1-x^{2} \leq 1$, and the value of $5 x-x^{2}$ is bounded by $5 x-x^{2}=\frac{25}{4}-\left(x-\frac{5}{2}\right)^{2} \leq \frac{25}{4}$, so

$$
\sqrt{1-x^{2}}+\sqrt{5 x-x^{2}} \leq \sqrt{1}+\sqrt{\frac{25}{4}}=1+\frac{5}{2}=3+\frac{1}{2}
$$

On the other hand, the value of the expression is non-negative and cannot be 0, as this would imply that $1-x^{2}=0$ and $5 x-x^{2}=0$, which is not possible. Thus, the only integer values that the expression can take are 1, 2, or 3.

Let $\sqrt{1-x^{2}}+\sqrt{5 x-x^{2}}=a$ and eliminate the roots. Instead of expressing the expression in this form, let's square $\sqrt{1-x^{2}}=a-\sqrt{5 x-x^{2}}$, which gives us $1-x^{2}=a^{2}-2 a \sqrt{5 x-x^{2}}+5 x-x^{2}$ or $2 a \sqrt{5 x-x^{2}}=a^{2}-1+5 x$. Squaring again, we get $4 a^{2}\left(5 x-x^{2}\right)=\left(a^{2}-1\right)^{2}-10 x+10 a^{2} x+25 x^{2}$ or

$$
x^{2}\left(25+4 a^{2}\right)+x\left(-10 a^{2}-10\right)+\left(a^{2}-1\right)^{2}=0
$$

If $a=1$, we get $x(29 x-20)=0$, from which it follows that $x=0$ or $x=\frac{20}{29}$. Substituting these solutions into the original expression, we find that only $x=0$ is valid.

For $a=2$, the quadratic equation for $x$ can be factored as $(x-1)(41 x-9)=0$, giving us $x=1$ or $x=\frac{9}{41}$. In both cases, the calculation shows that the value of the expression is indeed 2.

Finally, for $a=3$, we get the quadratic equation $61 x^{2}-100 x+64=0$. Since its discriminant $D=100^{2}-4 \cdot 61 \cdot 64=100^{2}-(2 \cdot 64)(2 \cdot 61)=100^{2}-128 \cdot 122$ is negative, it has no real solutions.

Thus, the value of the expression is an integer when $x=0$, $x=\frac{9}{41}$, or $x=1$.

The constraint $0 \leq a \leq 3$ 2 points
Writing the equation $x^{2}\left(25+4 a^{2}\right)+x\left(-10 a^{2}-10\right)+\left(a^{2}-1\right)^{2}=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1 point
Handling the cases $a \in\{0,1,2,3\}$ and excluding invalid $x \ldots \ldots \ldots$ 1 point each (If the validity of all $x$ is not checked in a particular case, the point is not awarded.)