1.

Find the value of k if the Quarditic equatation(k+1)^2×x -2(K-1)x+1=0 has equal roots.​

Answer»

ANSWER:

Correct OPTION is

A

k=0

C

k=3

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.


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