Q1. [1] § 2.2 Geometrical Meaning of the Zeroes of a Polynomial § 2.3 Relationship between Zeroes and Coefficients of a Polynomial
If the sum of zeroes of the polynomial $p(x) = 2x^2 - k\sqrt{2}\,x + 1$ is $\sqrt{2}$, then value of $k$ is :
- (a) $\sqrt{2}$
- (b) $2$
- (c) $2\sqrt{2}$
- (d) $\frac{1}{2}$
Previously asked in CBSE board exam
2024 30/1/1 Q1
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
(b) 2
For $p(x) = 2x^2 - k\sqrt{2}\,x + 1$, sum of zeroes $= \dfrac{k\sqrt{2}}{2} = \sqrt{2}$, so $k\sqrt{2} = 2\sqrt{2}$, giving $k = 2$.
Source: Chapter 2, Section 2.3
---
Explanation
Use the formula: sum of zeroes $= \dfrac{-b}{a}$. Here $b = -k\sqrt{2}$ and $a = 2$, so sum $= \dfrac{k\sqrt{2}}{2}$. Set this equal to $\sqrt{2}$ and solve for $k$. Examiners expect you to recall and apply the sum-of-zeroes formula correctly — no need to actually find the zeroes.
If a question refers to an image, map, graph or diagram that is not shown here, find the actual CBSE question paper on the CBSE website:
cbse.gov.in.