If one zero of the quadratic polynomial $(k - 1) x^{2} + k x + 1$ is -4 then the value of k is

(a) $\frac{- 5}{4}$ (b) $\frac{5}{4}$ (c) $\frac{- 4}{3}$ (d) $\frac{4}{3}$

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Solution

If one zero of the quadratic polynomial $p (x) = (k - 1) x^{2} + k x + 1 i s - 4$
then put $x = - 4$

$p (- 4) = (k - 1) (- 4)^{2} + k (- 4) + 1 = 0$

$= (k - 1) 16 - 4 k + 1 = 0$
$= 16 k - 16 - 4 k + 1 = 0$
$= 12 k - 15 = 0$
$= k = (\frac{15}{12}) = k = (\frac{5}{4})$

option (b) $(\frac{5}{4})$ is correct answer

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