Question

If one of the zeroes of the quadratic polynomial (k–1) x2 + k x + 1 is –3, then the value of k is

Answer 1

(k–1) x2 + k x + 1
${(k - 1)} { (- 3)}^{2} + k (- 3) + 1 \\ (k - 1) \times 9 - 3k + 1 = 0 \\ 9k - 9 - 3k + 1 = 0 \\ 6k - 8 = 0 \\ 6k = 8 \\ k = \frac{8}{6} \\ k = \frac{4}{3}$
is my answer

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 .