Question

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

Answer 1

given
quadratic polynomial = (k-1) x²+k x+1
one of the zeroes = -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
⇒ k = 8/6 = 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 .