Question

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

Answer 1

Answer:

Here k=$\frac{2}{3}$

Step-by-step explanation:

Given p(x)=(k-1)x²+kx+1

-3 is  a zero of the polynomial p(x)

⇒p(-3)=0

⇒(k-1)3²+k(3)+1=0

⇒9k-9+3k+1=0

⇒12k-8=0

⇒12k=8

⇒3k=2

⇒k=$\frac{2}{3}$

Answer 2

ANSWER

$f(x) = (k - 1)x² + kx + 1$

$f(3) = 0$

$(k - 1) × (3)² + (-3)k + 1 = 0$

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

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

$6k - 8 = 0$

$6k = 8$

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

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