Question

it one of the zeroes of quadratic polynomial (k-1)whole square+kx+1 is -3 then what is the value of K

Answer 1

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Hey mate !!
Here's the answer!!

Equation ->  (k-1)² + kx +1  = 0

One of the zeros is -3.
Substitute the value of x = -3 in the Equation.

(k-1)² + k*-3 +1 = 0
k² - 2(k)(1) +1 -3k = -1
k² - 2k -3k +1 = -1
k² - 5k + 1+1 = 0   = >  k² - 5k +2 = 0
k²  + k = -2
k(k-5) = -2

=> k = -2,
  k -5 = -2 = k =3

Therefore values of k are -2 and 3.

Hope it helps!!
Cheers mate!!

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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 .