Q1. [1] § 2.3 Relationship between Zeroes and Coefficients of a Polynomial § 2.4 Summary
If $\alpha$, $\beta$ are the zeroes of a polynomial $p(x) = x^2 + x - 1$, then $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ equals to
- A 1
- B 2
- C $-1$
- D $\dfrac{-1}{2}$
Previously asked in CBSE board exam
2023 30/1/1 Q7
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
Option A: 1
For $p(x) = x^2 + x - 1$: $\alpha + \beta = -1$, $\alpha\beta = -1$.
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{-1} = 1$$
Explanation
The key step is rewriting $\frac{1}{\alpha}+\frac{1}{\beta}$ as $\frac{\alpha+\beta}{\alpha\beta}$, then applying Vieta's formulas: sum $= \frac{-b}{a} = -1$ and product $= \frac{c}{a} = -1$. Dividing gives 1. Don't try to find individual zeroes — use the relations directly.
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