Question

x2 - 2(k-1)x + 1 has equal roots, find value of k.​

Answer 1

(k+1)x2+2(1−k)x+1=(ax+b)2  

and we are asked to find k. So:

a2=k+1

2ab=2(1−k)

b2=1,b=±1

±a=(1−k)

a2=(1−k)2=k+1

Solving (1−k)2=k+1 has k2−2k=k,k2=3k,k=0, k=3

Plugging in k=0, we have x2+2x+1=(x+1)2. Plugging in k=3, we have 4x2−4x+1=(1−2x)2

So the solutions are k=0,k=3

Answer 2

Answer:

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