Question

if one of the zeros of the quadratic polynomial (k-1)x^2 +kx+1 is -3, then the value of k is
(a)4/3 (b)-4/3 (c)2/3 (d)-2/3

Answer 1

Putting x=-3 in the equation,

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

9k-9-3k+1=0

6k-8=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 .