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

Options

A. 2
B. -2
C. 10
D. none

Answer

To solve the problem, we need to find the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(2x^2 + 6x + b = 0\) where \(b < 0\).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is \(2x^2 + 6x + b = 0\). Here, \(a = 2\), \(b = 6\), and \(c = b\).

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 \(\alpha \beta = \frac{c}{a}\)

For our equation:
- \(\alpha + \beta = -\frac{6}{2} = -3\)
- \(\alpha \beta = \frac{b}{2}\)

3. **Express \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)**:
We can rewrite \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) as:
\[ \frac{\alpha^2 + \beta^2}{\alpha \beta} \]

4. **Calculate \(\alpha^2 + \beta^2\)**:
We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we have:
\[ \alpha^2 + \beta^2 = (-3)^2 - 2 \left(\frac{b}{2}\right) = 9 - b \]

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

6. **Analyze the expression for \(b < 0\)**:
Since \(b < 0\), we can analyze the expression \(\frac{18 - 2b}{b}\):
- The term \(18 - 2b\) will be positive because \(b\) is negative (making \(-2b\) positive).
- Thus, \(\frac{18 - 2b}{b}\) will be negative since \(b\) is negative.

7. **Determine the limit**:
As \(b\) approaches negative values, the expression \(\frac{18 - 2b}{b}\) will decrease without bound. Therefore, we can conclude that:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} < -2 \]

### Conclusion:
Thus, \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) is less than \(-2\).