Question

find the value of k for which the given equation has real and equal roots. (k+1)x²-2(k-1)x+1=0 ​

Answer 1

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