$\alpha$ and $\beta$ are the zeroes of the polynomial $5x^2 - 16x - 10$. Find the value of $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer
For $5x^2 - 16x - 10$, we have $a = 5,\ b = -16,\ c = -10$.
$$\alpha + \beta = \frac{-b}{a} = \frac{16}{5}, \qquad \alpha\beta = \frac{c}{a} = \frac{-10}{5} = -2$$
Now,
$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$$
$$= \frac{\left(\dfrac{16}{5}\right)^2 - 2(-2)}{-2} = \frac{\dfrac{256}{25} + 4}{-2} = \frac{\dfrac{356}{25}}{-2} = \frac{-178}{25}$$
Source: Chapter 2, Section 2.3
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Explanation
- Examiners expect you to use the formulae $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ directly — do not find the zeroes individually.
- The key identity $\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$ must be stated explicitly for full credit.
- Arithmetic errors (especially with the negative product) are the most common source of mark loss here.