Question

If one of the zeros of the quadratic polynomial (k-1 x*2+kx+1 os -3 find value of k)

Answer 1

k-1x*2+kx+1=0
k-1(-3)*2+k*(-3)+1=0
k+3*2-3k+1=0
k+6-3k+1=0
-2k=  -7
k=7/2

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 .