Question

(k-x)x^ + k*x + 1 is -3 then find the value of k
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Answer 1

Answer:

P(x) $= (k - x)x^{2} + k*x + 1$

x = -3

P(x) = 0

$(k - x)x^{2} + k*x + 1 = 0$

$[k - (-3)](-3)^{2} + k*(-3) + 1 = 0$

$[k + 3](9) - 3k + 1 = 0$

$9k + 27 - 3k + 1 = 0$

$6k + 28 = 0$

$6k = - 28$

$k = \frac{-28}{6}$

$k = \frac{-14}{3}$