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

For $p(x) = x^2 - 3x - 1$, comparing with $ax^2 + bx + c$: $a = 1,\ b = -3,\ c = -1$.

$$\alpha + \beta = \frac{-b}{a} = \frac{3}{1} = 3, \qquad \alpha\beta = \frac{c}{a} = \frac{-1}{1} = -1$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{-1} = -3$$

Source: Chapter 2, Section 2.3

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• The key step is converting $\frac{1}{\alpha}+\frac{1}{\beta}$ into $\frac{\alpha+\beta}{\alpha\beta}$ — examiners expect this manipulation to be shown explicitly.
• Never find the actual zeroes; use Vieta's formulas directly. This saves time and avoids errors with irrational roots.
• Both marks are typically split: 1 mark for correctly stating $\alpha+\beta$ and $\alpha\beta$, and 1 mark for the final correct value $-3$.