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: Recall the relationships. For a quadratic polynomial $f(x) = ax^2 + bx + c$, the zeroes $\alpha$ and $\beta$ satisfy:
\[\alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}.\]
Here:
\[a = 6, \, b = 11, \, c = -10 \implies \alpha + \beta = -\frac{11}{6}, \, \alpha \beta = -\frac{10}{6}.\]
Step 2: Simplify the expression:
\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta}.\]
Use $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$:
\[\alpha^2 + \beta^2 = \left(-\frac{11}{6}\right)^2 - 2\left(-\frac{10}{6}\right) = \frac{121}{36} + \frac{20}{6}.\]
Simplify:
\[\alpha^2 + \beta^2 = \frac{121}{36} + \frac{120}{36} = \frac{241}{36}.\]
Substitute into $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$:
\[\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{241}{36}}{-\frac{10}{6}} = \frac{241}{60}.\]
Correct Answer: $\frac{241}{60}$.