Q1. [1] § 2.2 Geometrical Meaning of the Zeroes of a Polynomial § 2.3 Relationship between Zeroes and Coefficients of a Polynomial § 2.4 Summary
If $\alpha$, $\beta$ are the zeroes of the polynomial $p(x) = 4x^2 - 3x - 7$, then $\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\right)$ is equal to :
- (a) $\dfrac{7}{3}$
- (b) $\dfrac{-7}{3}$
- (c) $\dfrac{3}{7}$
- (d) $\dfrac{-3}{7}$
Previously asked in CBSE board exam
2023 30/4/1 Q17
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
(d) $\dfrac{-3}{7}$
For $p(x) = 4x^2 - 3x - 7$: $\alpha + \beta = \dfrac{3}{4}$, $\alpha\beta = \dfrac{-7}{4}$.
$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3/4}{-7/4} = \frac{-3}{7}$$
Source: Chapter 2, Section 2.3
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Explanation
- Use the standard formulae: sum of zeroes $= -b/a$ and product of zeroes $= c/a$.
- Rewrite $\frac{1}{\alpha}+\frac{1}{\beta}$ as $\frac{\alpha+\beta}{\alpha\beta}$ — this is the key algebraic step examiners look for.
- Here $a=4,\ b=-3,\ c=-7$, giving the answer $-3/7$. Don't forget the negative sign.
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