If `2(x-4)le(1)/(2)(5-3x)` and x is an integer, what is the smallest possible value of `x^(2)`?

Options

A. `(1)/(4)`
B. `1`
C. `4`
D. `9`

Answer

To solve the inequality \(2(x - 4) \leq \frac{1}{2}(5 - 3x)\) and find the smallest possible value of \(x^2\) where \(x\) is an integer, we can follow these steps:

### Step 1: Simplify the Inequality
Start by distributing on the left side:
\[ 2x - 8 \leq \frac{1}{2}(5 - 3x) \]

### Step 2: Eliminate the Fraction
Multiply both sides of the inequality by 2 to eliminate the fraction:
\[ 2(2x - 8) \leq 5 - 3x \]
This simplifies to:
\[ 4x - 16 \leq 5 - 3x \]

### Step 3: Rearrange the Inequality
Now, let's move all terms involving \(x\) to one side and constant terms to the other side:
\[ 4x + 3x \leq 5 + 16 \]
This simplifies to:
\[ 7x \leq 21 \]

### Step 4: Solve for \(x\)
Now, divide both sides by 7:
\[ x \leq 3 \]

### Step 5: Find Integer Values of \(x\)
Since \(x\) is an integer, the possible integer values for \(x\) are \(3, 2, 1, 0, -1, -2, \ldots\)

### Step 6: Calculate \(x^2\)
Now, we need to find the smallest possible value of \(x^2\):
- If \(x = 3\), then \(x^2 = 9\)
- If \(x = 2\), then \(x^2 = 4\)
- If \(x = 1\), then \(x^2 = 1\)
- If \(x = 0\), then \(x^2 = 0\)
- If \(x = -1\), then \(x^2 = 1\)
- If \(x = -2\), then \(x^2 = 4\)
- If \(x = -3\), then \(x^2 = 9\)

The smallest value of \(x^2\) is \(0\) when \(x = 0\).

### Conclusion
The smallest possible value of \(x^2\) is:
\[ \boxed{0} \]