Question

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

Answer 1

Step-by-step explanation:

HEYA MATE!!!

HERE IS YOUR ANSWER............

(K-1)x^2 + Kx + 1 where 'x' = -3

(K-1)(-3)^2 + k(-3) + 1 = 0

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

9k - 9 - 3K + 1 = 0

6K - 8 = 0

K = 8/ 6

K = 4/3

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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 .