Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + 11x - 10$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

Answer 1

Step 1: Understanding the problem:
We are given the quadratic polynomial \( f(x) = 6x^2 + 11x - 10 \), and we are asked to find the value of \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \), where \( \alpha \) and \( \beta \) are the zeroes of the polynomial.

Step 2: Use Vieta's formulas:
From Vieta's formulas for the quadratic equation \( ax^2 + bx + c = 0 \), we know the following relationships between the zeroes \( \alpha \) and \( \beta \):
- The sum of the zeroes is \( \alpha + \beta = -\frac{b}{a} \) - The product of the zeroes is \( \alpha \beta = \frac{c}{a} \) For the given polynomial \( f(x) = 6x^2 + 11x - 10 \), we have: - \( a = 6 \) - \( b = 11 \) - \( c = -10 \) So, according to Vieta's formulas: - \( \alpha + \beta = -\frac{11}{6} \) - \( \alpha \beta = \frac{-10}{6} = -\frac{5}{3} \) We are tasked with finding \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \). Using the identity:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} \] we need to find \( \alpha^2 + \beta^2 \).

Step 3: Find \( \alpha^2 + \beta^2 \):
We use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \] Substitute the values \( \alpha + \beta = -\frac{11}{6} \) and \( \alpha \beta = -\frac{5}{3} \):
\[ \alpha^2 + \beta^2 = \left( -\frac{11}{6} \right)^2 - 2 \times \left( -\frac{5}{3} \right) \] \[ \alpha^2 + \beta^2 = \frac{121}{36} + \frac{10}{3} \] Convert \( \frac{10}{3} \) to a fraction with denominator 36:
\[ \frac{10}{3} = \frac{120}{36} \] Now add the two fractions:
\[ \alpha^2 + \beta^2 = \frac{121}{36} + \frac{120}{36} = \frac{241}{36} \]

Step 4: Calculate \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \):
Now, substitute \( \alpha^2 + \beta^2 = \frac{241}{36} \) and \( \alpha \beta = -\frac{5}{3} \) into the formula:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{\frac{241}{36}}{-\frac{5}{3}} \] Simplify the division of fractions:
\[ \frac{\frac{241}{36}}{-\frac{5}{3}} = \frac{241}{36} \times \frac{3}{-5} = \frac{241 \times 3}{36 \times -5} = \frac{723}{-180} \] Simplify the fraction:
\[ \frac{723}{-180} = -\frac{241}{60} \]

Conclusion:
The value of \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \) is \( -\frac{241}{60} \).