Question

If P(x)=(k-1)x^2+kx+1 and one of the zeroes is negative 3, find value of k

Answer 1

p(x)=(k-1)x²+kx+1
If -3 is one of the zeros of p(x).
Then,p(-3)=0
Putting x=-3
(k-1)(-3)²+k(-3)+1=0
(k-1)9-3k+1=0
9k-9-3k+1=0
6k=8
k=4/3

Answer 2

Step-by-step explanation:

GIVEN:-)

→ One zeros of quadratic polynomial = -3.

→ Quadratic polynomial = ( k - 1 )x² + kx + 1.

Solution:-

→ P(x) = ( k -1 )x² + kx + 1 = 0.

→ p(-3) = ( k - 1 )(-3)² + k(-3) + 1 = 0.

=> ( k - 1 ) × 9 -3k + 1 = 0.

=> 9k - 9 -3k + 1 = 0.

=> 6k - 8 = 0.

=> 6k = 8.

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

Hence, the value of ‘k’ is founded .