Question

If 1 of the zereos of the quadratic equation polynomial (k-1)x²+kx+1 is -3 then tje value of k is

Answer 1

(k-1)-3^2+k-3+1=0

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

9k-9-3k+1=0

6k-9+1=0

6k=9-1

6k=8

k=8/6

k=4/3

Answer 2

Answer:

Step-by-step explanation:

(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