Question

one of the zeros of a quadratic polynomial (k+ 1)x square + kx + 1 is 2 find the value of k

Answer 1

Hey friend,

Here is your answer.
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p(x) = (k+1)x² + kx + 1

p(2) = (k+1)*2² + k*2 + 1 = 0

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

=> 4k + 4 + 2k + 1 = 0

=> 6k + 5 = 0

=> 6k = -5

=>  k  =  -5
               6
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Hope this helped you !!
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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 .