## 3. Solve the equation in the set of real numbers

$$
x\lfloor x\rfloor-5 x+7=0 \text {, }
$$

where $\lfloor x\rfloor$ denotes the floor function of $x$, i.e., the greatest integer $k$ such that $k \leqq x$. (For example, $\lfloor\sqrt{2}\rfloor=1 \quad \text{and} \quad \lfloor-3.1\rfloor=-4$).
SOLUTION. Let $k=\lfloor x\rfloor$, thus $x=k+\alpha, 0 \leqq \alpha<1$, and $k<5$. Solve the right inequality:

$$
\begin{aligned}
\frac{k^{2}-5 k+7}{5-k} & <1, \\
\frac{k^{2}-5 k+7-(5-k)}{5-k} & <0, \\
\frac{k^{2}-4 k+2}{5-k} & <0, \\
\frac{(k-2-\sqrt{2})(k-2+\sqrt{2})}{5-k} & <0 .
\end{aligned}
$$

According to the position of the numbers $2-\sqrt{2}, 2+\sqrt{2}$, and 5 on the number line, we find that the last inequality holds precisely when $k \in(2-\sqrt{2}, 2+\sqrt{2}) \cup(5, \infty)$. The inequalities $(*)$ therefore hold simultaneously precisely when $k \in(2-\sqrt{2}, 2+\sqrt{2})$. This condition is satisfied only by the three integers $k \in\{1,2,3\}$. For $k=1$, we compute $\alpha=\frac{3}{4}$, for $k=2$ we get $\alpha=\frac{1}{3}$, and for $k=3$ we have $\alpha=\frac{1}{2}$. In total, we obtain three solutions $x_{1}=\frac{7}{4}, x_{2}=\frac{7}{3}, x_{3}=\frac{7}{2}$.