Solve the following quantitative comparison problem by plugging in the number 0,1,2, -2, and `1//2` in that order - when possible.
`{:("Column A",x>0,"Column B"),(x^2+2,,x^3-2):}`

Options

A. If column A is larger
B. If column B is larger
C. If the columns are equal
D. If there is not enough information to decide

Answer

To solve the quantitative comparison problem, we will evaluate the expressions in Column A and Column B by plugging in the specified values of \( x \): 0, 1, 2, -2, and \( \frac{1}{2} \). However, we will only consider values of \( x \) that are greater than 0, as per the problem's constraint.

### Step-by-Step Solution:

1. **Plug in \( x = 1 \):**
- Column A: \( x^2 + 2 = 1^2 + 2 = 1 + 2 = 3 \)
- Column B: \( x^3 - 2 = 1^3 - 2 = 1 - 2 = -1 \)
- **Comparison:** Column A (3) is greater than Column B (-1).

2. **Plug in \( x = 2 \):**
- Column A: \( x^2 + 2 = 2^2 + 2 = 4 + 2 = 6 \)
- Column B: \( x^3 - 2 = 2^3 - 2 = 8 - 2 = 6 \)
- **Comparison:** Column A (6) is equal to Column B (6).

3. **Plug in \( x = \frac{1}{2} \):**
- Column A: \( x^2 + 2 = \left(\frac{1}{2}\right)^2 + 2 = \frac{1}{4} + 2 = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} \)
- Column B: \( x^3 - 2 = \left(\frac{1}{2}\right)^3 - 2 = \frac{1}{8} - 2 = \frac{1}{8} - \frac{16}{8} = -\frac{15}{8} \)
- **Comparison:** Column A (\(\frac{9}{4}\)) is greater than Column B (-\(\frac{15}{8}\)).

### Summary of Comparisons:
- For \( x = 1 \): Column A > Column B.
- For \( x = 2 \): Column A = Column B.
- For \( x = \frac{1}{2} \): Column A > Column B.

### Conclusion:
Since we have different results depending on the value of \( x \) (Column A is greater for \( x = 1 \) and \( x = \frac{1}{2} \), and equal for \( x = 2 \)), we conclude that there is not enough information to decide which column is consistently larger. Therefore, the correct answer is:

**D: There is not enough information to decide.**