find the value of k for which the given equation has real and equal roots. (k+1)x²-2(k-1)x+1=0
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Answer:
I hope it helps
Step-by-step explanation:
Given equation is (k+1)x
2
−2(k−1)x+1=0
This equation has real and equal roots
⟹Discriminant =0
⟹[−2(k−1)]
2
−4(k+1)(1)=0
⟹(k−1)
2
=(k+1)
⟹k
2
−2k+1=k+1
⟹k
2
=3k
⟹k=0 or 3
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