Solve the following equation (where `[*]` denotes greatest integer function and `{*}` represent fractional part function ) `[x^(2)]+2[x]=3x`, `0 le x le 2` .

Answer

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

### Step 1: Understand the Functions
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\), and the fractional part function \(\{x\} = x - [x]\) gives the non-integer part of \(x\).

### Step 2: Rewrite the Equation
We can express \(x\) as:
\[ x = [x] + \{x\} \]
Substituting this into the equation gives:
\[ [x^2] + 2[x] = 3([x] + \{x\}) \]

### Step 3: Analyze the Range of \(x\)
Since \(0 \leq x \leq 2\), we can consider the possible integer values of \([x]\):
- If \(0 \leq x < 1\), then \([x] = 0\)
- If \(1 \leq x < 2\), then \([x] = 1\)
- If \(x = 2\), then \([x] = 2\)

### Step 4: Case 1: \(0 \leq x < 1\)
Here, \([x] = 0\):
\[ [x^2] + 2(0) = 3([0] + \{x\}) \implies [x^2] = 3\{x\} \]
Since \(0 \leq x < 1\), \(\{x\} = x\) and \([x^2] = 0\) (as \(x^2 < 1\)):
\[ 0 = 3x \implies x = 0 \]
Thus, \(x = 0\) is a solution.

### Step 5: Case 2: \(1 \leq x < 2\)
Here, \([x] = 1\):
\[ [x^2] + 2(1) = 3(1 + \{x\}) \implies [x^2] + 2 = 3 + 3\{x\} \]
This simplifies to:
\[ [x^2] = 1 + 3\{x\} \]
Since \(1 \leq x < 2\), we have \(1 \leq x^2 < 4\). Thus, \([x^2]\) can be either 1, 2, or 3.

#### Subcase 2.1: If \([x^2] = 1\)
\[ 1 = 1 + 3\{x\} \implies 3\{x\} = 0 \implies \{x\} = 0 \implies x = 1 \]
Thus, \(x = 1\) is a solution.

#### Subcase 2.2: If \([x^2] = 2\)
\[ 2 = 1 + 3\{x\} \implies 3\{x\} = 1 \implies \{x\} = \frac{1}{3} \implies x = 1 + \frac{1}{3} = \frac{4}{3} \]
This value is valid since \(\frac{4}{3} < 2\).

#### Subcase 2.3: If \([x^2] = 3\)
\[ 3 = 1 + 3\{x\} \implies 3\{x\} = 2 \implies \{x\} = \frac{2}{3} \implies x = 1 + \frac{2}{3} = \frac{5}{3} \]
This value is valid since \(\frac{5}{3} < 2\).

### Step 6: Check the Values
The valid solutions we have found are:
- \(x = 0\)
- \(x = 1\)
- \(x = \frac{4}{3}\)
- \(x = \frac{5}{3}\)

### Step 7: Check \(x = 2\)
For \(x = 2\):
\[ [x^2] + 2[x] = [4] + 2[2] = 4 + 4 = 8 \]
\[ 3x = 3 \times 2 = 6 \]
This does not satisfy the equation.

### Final Solutions
Thus, the solutions to the equation \([x^2] + 2[x] = 3x\) for \(0 \leq x \leq 2\) are:
\[ x = 0, \quad x = 1, \quad x = \frac{4}{3}, \quad x = \frac{5}{3} \]