Question

If one of the zeroes of the quadratic polynomial (k-1)x²+kx+1 is -3 then find k

Answer 1

given that one of its zero is -3 so substitute -3 in given p(x)

SOLUTION:

P(x)=(k-1)x2+kx+1

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

k=8/6

k=4/3

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 .