2. Solve in the set of real numbers the equation: $x^{2}+2[x] \cdot\{x\}=3\left(3-\{x\}^{2}\right)$, where $[x]$ represents the integer part of $x$, and $\{x\}$ represents the fractional part of $x$.

Prof. Tamara Brutaru, Suceava
Solution. Since $\{x\}=x-[x]$, denoting $[x]=k, k \in \mathbb{Z}$, the equation becomes: $x^{2}+2 k(x-k)+3(x-k)^{2}=9$.

Performing the calculations, we obtain: $4 x^{2}-4 x k+k^{2}=9 \Leftrightarrow(2 x-k)^{2}=9$. It follows that $2 x-k=-3$ or $2 x-k=3$.

Since $2 x-k=2 x-[x]=2([x]+\{x\})-[x]=[x]+2\{x\}$, we have: $[x]+2\{x\}=-3$ or $[x]+2\{x\}=3$.

From $[x]+2\{x\}=-3$ and $[x] \in \mathbb{Z} \Rightarrow\{x\} \in\{0 ; 0.5\} \Rightarrow x \in\{-3 ;-3.5\}$, and from $[x]+2\{x\}=3$ and $[x] \in \mathbb{Z} \Rightarrow$ $\{x\} \in\{0 ; 0.5\} \Rightarrow x \in\{3 ; 2.5\}$

## Grading.

| $\{x\}=x-[x]$, for any real number $x$. | $\mathbf{1 p}$ |
| :--- | :--- |
| Denoting $[x]=k, k \in \mathbb{Z}$, the equation becomes: $x^{2}+2 k(x-k)+3(x-k)^{2}=9$. | $\mathbf{1 p}$ |
| Performing the calculations, we obtain: $4 x^{2}-4 x k+k^{2}=9 \Leftrightarrow(2 x-k)^{2}=9$. | $\mathbf{1 p}$ |
| It follows that $2 x-k=-3$ or $2 x-k=3$. | $\mathbf{1 p}$ |
| Since $2 x-k=2 x-[x]=2([x]+\{x\})-[x]=[x]+2\{x\}$, we have: $[x]+2\{x\}=-3$ or $[x]+2\{x\}=3$. | $\mathbf{1 p}$ |
| $[x]+2\{x\}=-3$ and $[x] \in \mathbb{Z} \Rightarrow\{x\} \in\{0 ; 0.5\} \Rightarrow x \in\{-3 ;-3.5\}$ | $\mathbf{1 p}$ |
| $[x]+2\{x\}=3$ and $[x] \in \mathbb{Z} \Rightarrow\{x\} \in\{0 ; 0.5\} \Rightarrow x \in\{3 ; 2.5\}$ | $\mathbf{1 p}$ |