if one of the zeros of quadratic polynomial (k-1)x2+kx+1 is -3then find the value of k
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Answered by
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...
-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
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. ...
Hence, the value of ‘k’ is founded .
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