Question

If one zero of the polynomial kx²+3x+k is 2 then find the value of k

Answer 1

Step-by-step explanation:

p(x) = kx^2 + 3x + k

p(2) = 0

[substitute x = 2]

k*2^2 + 3*3 + k = 0

4k + 6 + k = 0

5k = -6

k = -6/5 = -1.2

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 .