Question

if the zero of qudratic polynomial (k-1)x^2+kx+1 is 2​

Answer 1

$\boxed{SOLUTION \: by \: RKRAUSHAN}$

(K-1)x² +kx +1 = 0

put the value of x

(k-1)4 + 2k +1= 0

4k -4 +2k+1=0

6k =3

k= 3/6

$\huge \underline \red {Hope \: it \: helps} \\ \huge \underline \green{Thank \: U}$

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 .