Mathematics — Standard (041) — CBSE Class 10 board question

For $p(x) = x^2 + x - 2$, comparing with $ax^2 + bx + c$: $a = 1,\ b = 1,\ c = -2$.

Using the relations between zeroes and coefficients:
$$\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1$$
$$\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2$$

Now,
$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-1)^2 - 2(-2) = 1 + 4 = 5$$

$$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{5}{-2} = \mathbf{-\dfrac{5}{2}}$$

Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial

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• Examiners expect you to not find the actual zeroes but use the sum/product formulae directly — that's the point of this topic.
• The key identity is $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$; writing this step earns the middle mark.
• In a 3-mark question: 1 mark for writing sum and product, 1 mark for the identity/simplification, 1 mark for the correct final answer.