What is the value of x for which x[x] = 32 ?

Answer

To solve the equation \( x[x] = 32 \), where \( [x] \) denotes the greatest integer less than or equal to \( x \) (also known as the floor function), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Equation**:
We start with the equation \( x[x] = 32 \). Here, \( [x] \) is the greatest integer less than or equal to \( x \).

2. **Identifying Possible Values**:
Since \( x[x] = 32 \), we can express \( [x] \) as \( n \), where \( n \) is an integer. Thus, we can rewrite the equation as:
\[ x \cdot n = 32 \]
This implies:
\[ x = \frac{32}{n} \]

3. **Finding the Range for \( n \)**:
Since \( n = [x] \), it must satisfy \( n \leq x < n + 1 \). Substituting \( x \) from the previous step, we get:
\[ n \leq \frac{32}{n} < n + 1 \]

4. **Solving the Inequalities**:
- From the left inequality \( n \leq \frac{32}{n} \):
\[ n^2 \leq 32 \implies n \leq \sqrt{32} \approx 5.66 \]
Since \( n \) is an integer, the possible values for \( n \) are \( 1, 2, 3, 4, 5 \).

- From the right inequality \( \frac{32}{n} < n + 1 \):
\[ 32 < n(n + 1) \]
This can be rearranged to:
\[ n^2 + n - 32 > 0 \]
We can find the roots of the equation \( n^2 + n - 32 = 0 \) using the quadratic formula:
\[ n = \frac{-1 \pm \sqrt{1 + 128}}{2} = \frac{-1 \pm 11}{2} \]
This gives us \( n = 5 \) and \( n = -6 \). Since \( n \) must be positive, we have \( n > 5 \).

5. **Determining Valid \( n \)**:
The only integer value that satisfies both inequalities is \( n = 5 \).

6. **Calculating \( x \)**:
Now substituting \( n = 5 \) back into our equation for \( x \):
\[ x = \frac{32}{5} = 6.4 \]

7. **Final Verification**:
We check if \( [x] = 5 \):
\[ [6.4] = 5 \]
Thus, \( x[x] = 6.4 \cdot 5 = 32 \) confirms our solution.

### Conclusion:
The value of \( x \) for which \( x[x] = 32 \) is:
\[ \boxed{6.4} \]