If `m` and `n` are the zeros of the polynomial `3x^(2)+11x-4`, find the value of `(m)/(n)+(n)/(m)`.

Answer

To solve the problem, we need to find the value of \(\frac{m}{n} + \frac{n}{m}\) where \(m\) and \(n\) are the zeros of the polynomial \(3x^2 + 11x - 4\).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The polynomial is in the form \(ax^2 + bx + c\). Here, \(a = 3\), \(b = 11\), and \(c = -4\).

2. **Use the relationships of zeros**:
For a quadratic polynomial, the sum of the roots \(m + n\) and the product of the roots \(mn\) can be calculated using the formulas:
\[ m + n = -\frac{b}{a} \quad \text{and} \quad mn = \frac{c}{a} \]
Substituting the values:
\[ m + n = -\frac{11}{3} \quad \text{and} \quad mn = \frac{-4}{3} \]

3. **Express \(\frac{m}{n} + \frac{n}{m}\)**:
We can rewrite \(\frac{m}{n} + \frac{n}{m}\) as:
\[ \frac{m}{n} + \frac{n}{m} = \frac{m^2 + n^2}{mn} \]

4. **Find \(m^2 + n^2\)**:
We can use the identity:
\[ m^2 + n^2 = (m + n)^2 - 2mn \]
Substituting the values we found:
\[ m^2 + n^2 = \left(-\frac{11}{3}\right)^2 - 2\left(\frac{-4}{3}\right) \]
Calculating:
\[ m^2 + n^2 = \frac{121}{9} + \frac{8}{3} \]
To add these fractions, convert \(\frac{8}{3}\) to have a denominator of 9:
\[ \frac{8}{3} = \frac{24}{9} \]
Therefore:
\[ m^2 + n^2 = \frac{121}{9} + \frac{24}{9} = \frac{145}{9} \]

5. **Substitute back to find \(\frac{m^2 + n^2}{mn}\)**:
Now we substitute \(m^2 + n^2\) and \(mn\) into our expression:
\[ \frac{m^2 + n^2}{mn} = \frac{\frac{145}{9}}{\frac{-4}{3}} \]
This can be simplified by multiplying by the reciprocal:
\[ = \frac{145}{9} \times \frac{-3}{4} = \frac{-435}{36} \]

6. **Final answer**:
Thus, the value of \(\frac{m}{n} + \frac{n}{m}\) is:
\[ \frac{m}{n} + \frac{n}{m} = -\frac{145}{12} \]