If `b and c` are odd integers, then the equation `x^2 + bx + c = 0` has-

Options

A. two odd roots
B. two integer roots, one odd and one even
C. no integer roots
D. none of these

Answer

To solve the problem, we need to analyze the quadratic equation given by \(x^2 + bx + c = 0\), where \(b\) and \(c\) are odd integers. We will determine the nature of the roots of this equation.

### Step 1: Identify the roots using Vieta's formulas
Let \(\alpha\) and \(\beta\) be the roots of the equation. According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -b\)
- The product of the roots \(\alpha \beta = c\)

### Step 2: Analyze the sum of the roots
Since \(b\) is an odd integer, \(-b\) is also odd. Therefore:
\[ \alpha + \beta = -b \quad \text{(odd)} \]
This means that the sum of the roots is odd.

### Step 3: Analyze the product of the roots
Since \(c\) is an odd integer, we have:
\[ \alpha \beta = c \quad \text{(odd)} \]
This means that the product of the roots is odd.

### Step 4: Determine the nature of the roots
1. **Both roots odd**: If both roots \(\alpha\) and \(\beta\) were odd, then their sum \(\alpha + \beta\) would be even (since odd + odd = even). This contradicts our finding that \(\alpha + \beta\) is odd. Therefore, both roots cannot be odd.

2. **One root odd and one root even**: If one root is odd and the other is even, then their sum would still be odd (odd + even = odd). However, the product of an odd and an even number is even, which contradicts our finding that \(\alpha \beta\) is odd. Thus, this scenario is also not possible.

3. **No integer roots**: Given that both the sum and product of the roots are odd, it suggests that the roots cannot be integers. For the roots to be non-integers, the discriminant must not be a perfect square. The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
Substituting \(a = 1\) and \(c\) (which is odd), we see that \(D\) is unlikely to be a perfect square, confirming that the roots are not integers.

### Conclusion
Since both scenarios of having both roots odd or one root odd and one root even lead to contradictions, we conclude that the equation \(x^2 + bx + c = 0\) has **no integer roots**.