Question

If one of the zeroes of the quadratic polynomial (k-1)x^+kx+1is-3 find k

Answer 1

Let the given quadratic polynomial be p(x) = (k-1)x2 +kx+1. It is given that one of its zeros is -3.

Therefore, p(-3) = 0

⇒ (k-1)(-3)2 +k(-3)+1 = 0

⇒ 9k-9-3k+1 = 0

⇒ 6k - 8 = 0

⇒ k =

4/3
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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 .