Question

If one of the zero of the quadratic polynomial k - 1 x square + kx + 1 is minus 3 then the value of k is

Answer 1

Answer:

ITS VERY SIMPLE

WE HAVE GIVEN VALUE OF X = -3

SO, (K-1)X² + KX + 1 = 0

==> (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 .