If one of the zeroes of a quadratic polynomial $(k - 1) x^{2} + k x + 1$ is -3, then find the value of k.

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Solution

Given $p (x) = (k - 1) x^{2} + k x + 1$ and $p (- 3) = 0$
$∴ (k - 1) (- 3)^{2} + k (- 3) + 1 = 0$
$\Rightarrow 9 (k - 1) - 3 k + 1 = 0$
$\Rightarrow 9 k - 9 - 3 k + 1 = 0$
$\Rightarrow 6 k - 8 = 0$
$\Rightarrow 6 k = 8$
$\Rightarrow k = \frac{8}{6}$
$\Rightarrow k = \frac{4}{3}$

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Similar questions

(k - 1) x^{2} + k x + 1

(k - 1) x^{2} + k x + 1

(A) \frac{4}{3}

(B) \frac{- 4}{3}

(C) \frac{2}{3}

(D) \frac{- 2}{3}

\frac{- 5}{4}

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If one of the zeros of quadratic polynomial (k - 1) x^{2} + k x + 1 is - 3,

then find the value of k .

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