If `alpha` and `beta` are the zeroes of the polynomial f(x) = `5x^(2) - 7x + 1` then find the value of `((alpha)/(beta)+(beta)/(alpha))` .

Answer

To find the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) where \(\alpha\) and \(\beta\) are the zeros of the polynomial \(f(x) = 5x^2 - 7x + 1\), we can follow these steps:

### Step 1: Identify the coefficients
The polynomial is in the form \(ax^2 + bx + c\), where:
- \(a = 5\)
- \(b = -7\)
- \(c = 1\)

### Step 2: Calculate the sum and product of the zeros
Using the formulas for the sum and product of the zeros:
- Sum of the zeros \((\alpha + \beta) = -\frac{b}{a} = -\frac{-7}{5} = \frac{7}{5}\)
- Product of the zeros \((\alpha \beta) = \frac{c}{a} = \frac{1}{5}\)

### Step 3: Use the identity for \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
We know that:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]

### Step 4: Find \(\alpha^2 + \beta^2\)
We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we calculated:
\[ \alpha^2 + \beta^2 = \left(\frac{7}{5}\right)^2 - 2 \cdot \frac{1}{5} \]
Calculating \(\left(\frac{7}{5}\right)^2\):
\[ \left(\frac{7}{5}\right)^2 = \frac{49}{25} \]
Now, calculate \(2 \cdot \frac{1}{5} = \frac{2}{5} = \frac{10}{25}\) (to have a common denominator):
\[ \alpha^2 + \beta^2 = \frac{49}{25} - \frac{10}{25} = \frac{39}{25} \]

### Step 5: Substitute back to find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
Now we substitute \(\alpha^2 + \beta^2\) and \(\alpha \beta\) into the equation:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{39}{25}}{\frac{1}{5}} = \frac{39}{25} \cdot 5 = \frac{39}{5} \]

### Final Answer
Thus, the value of \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) is \(\frac{39}{5}\).

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