If `alpha and beta` are the roots of equation `6x^(2)+x-2=0`, find the value of `(alpha)/(beta)+(beta)/(alpha)`:

Options

A. `-(25)/(12)`
B. `-(12)/(75)`
C. `-(4)/(9)`
D. none of these

Answer

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

### Step 1: Identify the coefficients of the quadratic equation
The given quadratic equation is in the form \(ax^2 + bx + c = 0\). Here, we have:
- \(a = 6\)
- \(b = 1\)
- \(c = -2\)

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{1}{6}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{-2}{6} = -\frac{1}{3}\)

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

### Step 4: Calculate \(\alpha^2 + \beta^2\)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = \left(-\frac{1}{6}\right)^2 - 2\left(-\frac{1}{3}\right) \]
Calculating:
\[ = \frac{1}{36} + \frac{2}{3} \]
To add these fractions, we convert \(\frac{2}{3}\) to have a common denominator of 36:
\[ \frac{2}{3} = \frac{2 \times 12}{3 \times 12} = \frac{24}{36} \]
Thus:
\[ \alpha^2 + \beta^2 = \frac{1}{36} + \frac{24}{36} = \frac{25}{36} \]

### Step 5: Substitute back to find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
Now we substitute \(\alpha^2 + \beta^2\) and \(\alpha \beta\) into our expression:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{25}{36}}{-\frac{1}{3}} \]
Calculating this gives:
\[ = \frac{25}{36} \times -3 = -\frac{25 \times 3}{36} = -\frac{75}{36} = -\frac{25}{12} \]

### Final Answer
Thus, the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) is:
\[ \boxed{-\frac{25}{12}} \]