Question

If one of the zeros of (k-1) x2 + kx +1 is -3 , then the value of k is​

Answer 1

Answer:

p(x) = (k-1) ×2 + kx +1

p(-3) = (k-1) ×2 +k×-3 +1

p(-3) = (k-1) ×2 -3k +1

p(-3) = 2k-2 -3k +1

p(-3) = 2k-3k-2+1

p(-3) = -k-1

-k-1 =0

-k = 1

k = -1