Question

If -3 is one of the zeroes of the polynomial (k-1)x^2 + kx+1, find the value of K?

Answer 1

$p(x) = (k - 1) {x}^{2} + kx + 1$

$p( - 3) = 0$

Substituting the value of p,

$(k - 1) { - 3}^{2} + k( - 3) + 1 = 0$

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

$= 6k - 8 = 0$

$= 6k = 8$

$= k = \frac{8}{6} = \frac{4}{3} \: or \: 1.3$

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 .