if-3is one of the zeroes of polynomial (k-1)x^2+kx+1 find the value of k
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Answered by
6
Heres your answer..
p(-3)= (k-1)(-3)^2 -3k + 1
= 9k-9-3k+1=0
= 6k= -1+9
= 6k= 8
k=8/6= 4/3
Hope it Helps
p(-3)= (k-1)(-3)^2 -3k + 1
= 9k-9-3k+1=0
= 6k= -1+9
= 6k= 8
k=8/6= 4/3
Hope it Helps
Answered by
0
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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