Solve for x if `(x)^(2) + (x+1)^(2) = 25`, where (x) denotes the least integer function

Options

A. `[-5, -4] cup [2, 3]`
B. `[-5, -4] cup [2, 3] cup [57]`
C. `[-5, -4] cup [2,3] cup {1}`
D. None of these

Answer

To solve the equation \( (x)^2 + (x+1)^2 = 25 \) where \( (x) \) denotes the least integer function, we can follow these steps:

### Step 1: Understand the least integer function
The least integer function, denoted as \( (x) \), gives the greatest integer less than or equal to \( x \). If \( x \) is an integer, then \( (x) = x \). If \( x \) is not an integer, then \( (x) \) is the integer part of \( x \).

### Step 2: Assume \( x \) is an integer
Let \( x = k \), where \( k \) is an integer. The equation becomes:
\[ k^2 + (k + 1)^2 = 25 \]

### Step 3: Expand the equation
Expanding \( (k + 1)^2 \):
\[ k^2 + (k^2 + 2k + 1) = 25 \]
This simplifies to:
\[ 2k^2 + 2k + 1 = 25 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 2k^2 + 2k + 1 - 25 = 0 \]
\[ 2k^2 + 2k - 24 = 0 \]
Dividing the entire equation by 2:
\[ k^2 + k - 12 = 0 \]

### Step 5: Factor the quadratic equation
We factor \( k^2 + k - 12 \):
\[ (k - 3)(k + 4) = 0 \]
This gives us the solutions:
\[ k = 3 \quad \text{or} \quad k = -4 \]

### Step 6: Determine values of \( x \)
Since \( k = (x) \), we have:
1. If \( k = 3 \), then \( x \) can be \( 3 \) (as it is an integer).
2. If \( k = -4 \), then \( x \) can be \( -4 \) (as it is an integer).

### Step 7: Consider non-integer values of \( x \)
Now, let’s consider \( x \) in the form \( x = k + f \), where \( k \) is an integer and \( 0 \leq f < 1 \). The equation becomes:
\[ (k + f)^2 + (k + f + 1)^2 = 25 \]
Expanding this:
\[ (k^2 + 2kf + f^2) + ((k + f + 1)^2) = 25 \]
This simplifies to:
\[ k^2 + 2kf + f^2 + (k^2 + 2k + 2kf + f^2 + 2f + 1) = 25 \]
Combining like terms:
\[ 2k^2 + 4kf + 2k + 2f^2 + 2f + 1 = 25 \]
Rearranging gives:
\[ 2k^2 + 4kf + 2k + 2f^2 + 2f - 24 = 0 \]
Dividing by 2:
\[ k^2 + 2kf + k + f^2 + f - 12 = 0 \]

### Step 8: Solve for \( k \)
To find integer values of \( k \), we can substitute \( f = 0 \) (since \( f \) is less than 1):
\[ k^2 + k - 12 = 0 \]
This gives us the same solutions \( k = 3 \) or \( k = -4 \).

### Step 9: Determine ranges for \( x \)
1. For \( k = 3 \): \( x = 3 + f \) where \( 0 \leq f < 1 \) gives \( x \) in the interval \( [3, 4) \).
2. For \( k = -4 \): \( x = -4 + f \) where \( 0 \leq f < 1 \) gives \( x \) in the interval \( [-4, -3) \).

### Final Solution
Thus, the solution set for \( x \) is:
\[ x \in [-4, -3) \cup [3, 4) \]