Solve the following equations
`x^(2)-3-{x}=0`
where [x] denotes the greatest integer and {x} fractional part.

Answer

To solve the equation \( x^2 - 3 - \{x\} = 0 \), where \([x]\) denotes the greatest integer function and \(\{x\}\) denotes the fractional part of \(x\), we can follow these steps:

### Step 1: Understand the relationship between \(x\), \([x]\), and \(\{x\}\)
The fractional part of \(x\) is defined as:
\[ \{x\} = x - [x] \]
This means that \(0 \leq \{x\} < 1\).

### Step 2: Rearrange the equation
We can rearrange the original equation:
\[ \{x\} = x^2 - 3 \]

### Step 3: Establish the bounds for \(\{x\}\)
Since \(\{x\}\) must be between 0 and 1, we can set up the following inequalities:
\[ 0 \leq x^2 - 3 < 1 \]

### Step 4: Solve the inequalities
**Inequality 1:**
\[ x^2 - 3 \geq 0 \implies x^2 \geq 3 \implies x \leq -\sqrt{3} \quad \text{or} \quad x \geq \sqrt{3} \]

**Inequality 2:**
\[ x^2 - 3 < 1 \implies x^2 < 4 \implies -2 < x < 2 \]

### Step 5: Combine the results
Now we combine the results from the two inequalities:
1. From \(x^2 \geq 3\), we have \(x \leq -\sqrt{3}\) or \(x \geq \sqrt{3}\).
2. From \(x^2 < 4\), we have \(-2 < x < 2\).

Thus, we need to find the intersection of these ranges:
- For \(x \geq \sqrt{3}\), we have \(\sqrt{3} \leq x < 2\).
- For \(x \leq -\sqrt{3}\), we have \(-2 < x \leq -\sqrt{3}\).

### Step 6: Write the final solution
The solution set can be expressed as:
\[ x \in [\sqrt{3}, 2) \cup (-2, -\sqrt{3}] \]