If x and y are non-zero real numbers, then `x^(2) + xy + y^(2)`

Options

A. is always positive
B. is always negative
C. takes the value zero for some x and y
D. takes both positive and negative values

Answer

To determine whether the expression \( x^2 + xy + y^2 \) is always positive, negative, or zero for non-zero real numbers \( x \) and \( y \), we can analyze the expression step by step.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression we are analyzing is \( x^2 + xy + y^2 \). This consists of three terms: \( x^2 \), \( xy \), and \( y^2 \).

2. **Identifying the Terms**:
- \( x^2 \) is the square of \( x \).
- \( y^2 \) is the square of \( y \).
- \( xy \) is the product of \( x \) and \( y \).

3. **Considering Non-Zero Values**:
Since \( x \) and \( y \) are non-zero real numbers, both \( x^2 \) and \( y^2 \) will always be positive (as the square of any non-zero real number is positive).

4. **Analyzing Different Cases**:
- **Case 1**: Both \( x \) and \( y \) are positive.
- Here, \( x^2 > 0 \), \( y^2 > 0 \), and \( xy > 0 \). Thus, \( x^2 + xy + y^2 > 0 + 0 + 0 = 0 \).
- **Case 2**: Both \( x \) and \( y \) are negative.
- Here, \( x^2 > 0 \), \( y^2 > 0 \), and \( xy > 0 \) (since the product of two negative numbers is positive). Thus, \( x^2 + xy + y^2 > 0 + 0 + 0 = 0 \).
- **Case 3**: One of \( x \) or \( y \) is positive and the other is negative.
- Without loss of generality, let \( x > 0 \) and \( y < 0 \). The term \( xy \) will be negative, but \( x^2 > 0 \) and \( y^2 > 0 \). The overall expression becomes \( x^2 + xy + y^2 \).
- For example, let \( x = 2 \) and \( y = -3 \):
\[ x^2 + xy + y^2 = 2^2 + (2)(-3) + (-3)^2 = 4 - 6 + 9 = 7 > 0. \]

5. **Conclusion**:
In all cases, whether both numbers are positive, both are negative, or one is positive and the other is negative, the expression \( x^2 + xy + y^2 \) is always positive.

### Final Result:
Thus, we conclude that \( x^2 + xy + y^2 > 0 \) for all non-zero real numbers \( x \) and \( y \).