Math, asked by faizuuu, 1 year ago

if one of the zeros of quadratic polynomial (k-1)x2+kx+1 is -3then find the value of k

Answers

Answered by snehitha2
12
(k-1)x²+kx+1=0

-3 is zero of the polynomial.

Put x= -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

k=8/6

k=4/3

Hope it helps...

Answered by Anonymous
1

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 .

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