Question

(k+1)x²-2(k+3)x+(2k+3)=0​

Answer 1

Answer:

The given equation is

(k+1)x

2

−2(k−1)x+1=0

comparing it with ax

2

+bx+c=0 we get

a=(k+1),b=−2(k−1) and c=1

∴ Discriminant,

D=b

2

−4ac=4(k−1)

2

−4(k+1)×1

=4(k

2

−2k+1)−4k−4

⇒4k

2

−8k+4−4k−4=4k

2

−12k

Since roots are real and equal, so

D=0⇒4k

2

−12k=0⇒4k(k−3)=0

⇒ either k=0 or k−3=0⇒4k(k−3)=0

Hence, k=0,3.