Question

If one of the zeroes of the quadratic polynomial (k – 1)x^2 + kx + 1 is -3, then the value of k is



above pic are o options​

Answer 1

1st option is the correct option

Answer 2

Step-by-step explanation:

$f(x) = (k - 1) {x}^{2} + kx + 1$

$f( - 3) = (k - 1) \times 9 + k( - 3) + 1 = 9k - 9 - 3k + 1 = 6k - 8$

If y is zero of polynomial

$a {x}^{2} + bx + c$

then f(y) = 0

$6k - 8 = 0$

$= > 6k = 8$

$= > k = \frac{8}{6} = \frac{4}{3}$

Answer is a) part

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