Question

If one of the zeros of the quadratic polynomial (k-1)x^2+kx+1 is -3 then find the value of k

Answer 1

put x=-3 in the given equation....
(k-1)(-3)^2+k(-3)+1=0
(k-1)9 -3k +1 =0
9k-9-3k+1=0
6k-8 = 0
2(3k-4) =0
3k-4=0
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 .