Question

If a value of polynomial f(x) = kx 2 + (3k-1)x +k is 3, find value of k.

Answer 1

f(3)=k(3)2+(3k-1)3+k
f(3)= 6k+9k-3+k
f(3)=16k-3
f(3)=0
16k-3=0
16k=3
k=3/16

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 .