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

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Solution

$Given: p (x) = (k - 3) x^{2} + k x + 3 and p (- 2) = 0$
Putting value of x equal to -2 in p(x) we get,

$∴ (k - 3) (- 2)^{2} + k (- 2) + 3 = 0 = 4 (k - 3) - 2 k + 3 = 0 = 4 k - 12 - 2 k + 3 = 0 = 2 k - 9 = 0 = 2 k = 9 \Rightarrow k = \frac{9}{2}$ .

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