Question

if one of the zero of polynomial (k-1)x(x)+kx+1 is -3 find the value of k

Answer 1

(k-1)(-3)(-3)+k(-3)+1=0 (K-1)9-3(k)+1=0 9k-9-3k+1=0 6k-8=0 6k=8 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 .