If the zeroes of $x^2 - k x + 6$ are in the ratio $2 : 3$, then $k$ is equal to:
(a) $\pm 5$
(b) $\pm 6$
(c) $\pm 1$
(d) $\pm 4$
Step-by-Step Solution
Key Concept: Let zeroes be $2r, 3r$. Product $= 6r^2 = 6 \Rightarrow r^2 = 1 \Rightarrow r = \pm 1$. Sum $= 5r = k \Rightarrow k = \pm 5$.
Zeroes $= 2r, 3r$. Product $= 6r^2 = 6 \Rightarrow r = \pm 1$. [0.5 Mark]
Sum $= 5r = k \Rightarrow k = \pm 5$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Solving $6r^2 = 6 \Rightarrow r = \pm 1$: 0.5 Mark
Finding $k = \pm 5$: 0.5 Mark
Correct Answer: $\pm 5$