If x, y, z are real and distinct, then `x^(2) + 4 y^(2) + x + 1 = 0`, is

Options

A. non-negative
B. non-positive
C. zero
D. none of these

Answer

To determine whether the expression \( x^2 + 4y^2 + x + 1 \) is non-negative, non-positive, zero, or none of these, we can analyze the expression step by step.

### Step-by-Step Solution:

1. **Start with the given expression**:
\[ x^2 + 4y^2 + x + 1 \]

2. **Rearrange the expression**:
We can group the terms involving \( x \):
\[ x^2 + x + 4y^2 + 1 \]

3. **Complete the square for the \( x \) terms**:
The expression \( x^2 + x \) can be rewritten by completing the square:
\[ x^2 + x = (x + \frac{1}{2})^2 - \frac{1}{4} \]
Therefore, we can rewrite the entire expression as:
\[ (x + \frac{1}{2})^2 - \frac{1}{4} + 4y^2 + 1 \]

4. **Combine constants**:
Now, combine the constants:
\[ (x + \frac{1}{2})^2 + 4y^2 + \frac{3}{4} \]

5. **Analyze the expression**:
The term \( (x + \frac{1}{2})^2 \) is always non-negative since it is a square. The term \( 4y^2 \) is also non-negative since it is a square multiplied by a positive constant. Thus, both terms are \( \geq 0 \).

6. **Conclusion**:
Therefore, the entire expression:
\[ (x + \frac{1}{2})^2 + 4y^2 + \frac{3}{4} \geq \frac{3}{4} \]
This means that the expression \( x^2 + 4y^2 + x + 1 \) is always greater than or equal to \( \frac{3}{4} \) and hence is **non-negative**.

### Final Result:
The expression \( x^2 + 4y^2 + x + 1 \) is **non-negative**.

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