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Q.

If one of the zeroes of the quadratic polynomial is $(k - 1) x^{2} + kx + 1$ is -3, then k equal to

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a

$\frac{2}{3}$

b

$\frac{4}{3}$

c

$- \frac{2}{3}$

d

$- \frac{4}{3}$

answer is A.

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Detailed Solution

$x = - 3 \begin{array}{l} p (x) = (K - 1) x^{2} + Kx + 1 \\ p (- 3) = 0 \\ (K - 1) (- 3)^{2} + K (- 3) + 1 = 0 \\ (K - 1) 9 - 3 K + 1 = 0 \\ 9 K - 9 - 3 K + 1 = 0 \\ 6 K - 8 = 0 \\ 6 K = 8 \\ K = \frac{8}{6} = \frac{4}{3} K = \frac{4}{3} \end{array}$

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