Subject (2). Solve in the set of real numbers the equation:

$$
x^{2}+2\lfloor x\rfloor\{x\}+3\{x\}^{2}=4
$$

G.M.-B. $n$. $6-7-8 / 2014$
Solution. We have, by elementary transformations

$4=x^{2}+2\lfloor x\rfloor\{x\}+3\{x\}^{2}=(\lfloor x\rfloor+\{x\})^{2}+2\lfloor x\rfloor\{x\}+3\{x\}^{2}=(\lfloor x\rfloor+2\{x\})^{2}$.

Therefore, $\lfloor x\rfloor+2\{x\}= \pm 2$, so $\{x\} \in\{0,1 / 2\}$, since $\lfloor x\rfloor \in \mathbb{Z}$ and $0 \leq\{x\}<1$. There are four solutions $x \in\{-5 / 2,-2,2,3 / 2\}$.