Question

if one of the zeroes of the quadratic polynomial (k-1)+k×x+1 is (-3),then find the value of k

Answer 1

Given one zero of quadratic polynomial (k-1)*2+kx+1 is - 3
So when we substitute - 3 in the polynomial, we get 0
(k-1)square of (-3) +k(-3)+1=0
9k-9-3k+1=0
6k-8=0
K=8/6 =4/3

I hope this will help you

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 .