Question

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

Answer 1

Answer:

Step-by-step explanation:

Given one of the zeros of quadratic polynomial (k+1)x²+kx+1 is -3

So when we substitute -3 in the polynomial, we get 0

→(k+1)(-3)²+k(-3)+1=0

→(k+1) (9) +k(-3)+1 = 0

→(9k+9)-3k+1=0

→6k+10=0

→k = -10/6 = -5/3

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 .