If one of the zeroes of the quadratic polynomial $(k-1)x^2 + kx + 1$ is $-3$, then the value of $k$ is:
(a) $\dfrac{4}{3}$
(b) $-\dfrac{4}{3}$
(c) $\dfrac{2}{3}$
(d) $-\dfrac{2}{3}$
Step-by-Step Solution
Key Concept: If $\alpha$ is a zero of polynomial $p(x)$, then $p(\alpha) = 0$.
Since $-3$ is a zero, $p(-3) = (k-1)(-3)^2 + k(-3) + 1 = 0$. [0.5 Mark]
$9(k-1) - 3k + 1 = 0 \Rightarrow 9k - 9 - 3k + 1 = 0 \Rightarrow 6k - 8 = 0 \Rightarrow k = \dfrac{8}{6} = \dfrac{4}{3}$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Substituting $x = -3$ into $p(x) = 0$: 0.5 Mark
Solving for $k$: 0.5 Mark
Correct Answer: $\dfrac{4}{3}$