Question

if one the zeroes of quadratic polynomial (k-1)x²+kx+1 is -3 . then the value of k is ?

Answer 1

X=-3(Given)

(k-1)x^2+ke+1=0
(k-1)(-3)^2+k(-3)+1=0
(k-1)(9)-3k+1=0
9k-9-3k+1=0
6k-8=0
6k=8
k=8/6
k=4/3

Answer 2

$f(x) = (k - 1)x² + kx + 1$

$f(3) = 0$

$(k - 1) × (3)² + (-3)k + 1 = 0$

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

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

$6k - 8 = 0$

$6k = 8$

$k = \frac{8}{6}$

$k = \frac{4}{3}$