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 \(\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)\) for the polynomial \(f(x) = 5x^2 - 7x + 1\), follow these steps:

1. **Identify the polynomial coefficients:**
The given polynomial is \(f(x) = 5x^2 - 7x + 1\).
- Coefficient \(a = 5\)
- Coefficient \(b = -7\)
- Coefficient \(c = 1\)

2. **Find the sum and product of the zeroes:**
- Sum of the zeroes \(\alpha + \beta = -\frac{b}{a}\)
- Product of the zeroes \(\alpha \beta = \frac{c}{a}\)

Using the given coefficients:
\[ \alpha + \beta = -\left(\frac{-7}{5}\right) = \frac{7}{5} \]
\[ \alpha \beta = \frac{1}{5} \]

3. **Express \(\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)\) in terms of \(\alpha\) and \(\beta\):**
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \]

4. **Simplify \(\alpha^2 + \beta^2\) using the identity:**
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]

Substitute the known values:
\[ \alpha^2 + \beta^2 = \left(\frac{7}{5}\right)^2 - 2 \left(\frac{1}{5}\right) \]
\[ \alpha^2 + \beta^2 = \frac{49}{25} - \frac{2}{5} \]

5. **Convert \(\frac{2}{5}\) to a common denominator:**
\[ \frac{2}{5} = \frac{2 \times 5}{5 \times 5} = \frac{10}{25} \]

6. **Subtract the fractions:**
\[ \alpha^2 + \beta^2 = \frac{49}{25} - \frac{10}{25} = \frac{39}{25} \]

7. **Divide by \(\alpha \beta\):**
\[ \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{39}{25}}{\frac{1}{5}} = \frac{39}{25} \times \frac{5}{1} = \frac{39 \times 5}{25} = \frac{195}{25} = 7.8 \]

Therefore, the value of \(\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)\) is \(7.8\).