If `alpha, beta` are roots of the equation `2x^2 + 6x + b = 0 (b < 0),` then `alpha/beta+beta/alpha` is less than

Options

A. 2
B. -2
C. 18
D. none of these

Explanation

To solve the problem, we need to analyze the quadratic equation given and find the expression \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \) in terms of the roots \( \alpha \) and \( \beta \).

### Step-by-step Solution:

1. **Identify the quadratic equation**:
The given equation is:
\[ 2x^2 + 6x + b = 0 \]

2. **Use Vieta's formulas**:
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Here, \( a = 2 \), \( b = 6 \), and \( c = b \). Therefore:
\[ \alpha + \beta = -\frac{6}{2} = -3 \]
\[ \alpha \beta = \frac{b}{2} \]

3. **Express \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \)**:
We can rewrite the expression using the identity:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]
To find \( \alpha^2 + \beta^2 \), we use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we have:
\[ \alpha^2 + \beta^2 = (-3)^2 - 2 \left(\frac{b}{2}\right) = 9 - b \]

4. **Substituting back into the expression**:
Now substituting \( \alpha^2 + \beta^2 \) and \( \alpha \beta \) into our expression:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{9 - b}{\frac{b}{2}} = \frac{2(9 - b)}{b} = \frac{18 - 2b}{b} \]

5. **Analyzing the expression**:
Since \( b < 0 \), we can analyze the expression:
\[ \frac{18 - 2b}{b} = \frac{18}{b} - 2 \]
As \( b \) approaches negative infinity, \( \frac{18}{b} \) approaches 0. Therefore:
\[ \frac{18}{b} - 2 \to -2 \text{ as } b \to -\infty \]

6. **Conclusion**:
Since \( b < 0 \), the value of \( \frac{18}{b} - 2 \) will always be less than -2. Thus:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} < -2 \]

### Final Answer:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \text{ is less than } -2. \]