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 :

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Accepted Answer

## 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.

Mathematics — CBSE Class 10 board question