Question

if one of the zeros of the quadratic polynomial ( k-1 ) x square + kx + 1 is -3 , then find the value of k

Answer 1

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 .

Answer 2

$f(x) = (k - 1)x² + kx + 1$

$f(3) = 0$

$(k - 1) × (3)² + (-3)k + 1 = 0$

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

$9k - 9 - 3k + 1 = 0$

$6k - 8 = 0$

$6k = 8$

$k = \frac{8}{6}$

$\boxed{k = \frac{4}{3}}$