Question

if one of the zeros of the quadratic , (k-1)x2 +kx+l is -3 then the value of k is

Answer 1

if zeros is -3 then on putting x = -3 equation becomes zero
(k-1)x^2+Kx+1=0 ( if x= -3)
(k-1)9-3k+1 = 0
9k-9-3k+1 = 0
6k-8= 0
k = 8/6
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 .