Question

Find the value of k for which the equation x 2 +k(2 x + k − 1) + 2 = 0 has real and equal roots.

Answer 1

Hey Mate
you're Answer --

The given equation is  x 2 + k 2 x + k - 1 + 2 = 0 . 
⇒ x 2 + 2 k x + k k - 1 + 2 = 0 
So,  a = 1,  b = 2 k, c =  k( k − 1) + 2 
We know  D = b 2 - 4 a c 
⇒ D = 2 k 2 - 4 × 1 × k k - 1 + 2 ⇒ D = 4 k 2 - 4 k2 - k + 2 ⇒ D = 4 k 2 - 4 k 2 + 4 k - 8 ⇒ D = 4 k - 8= 4 k - 2 
For equal roots,  D = 0 
Thus, 4( k − 2) = 0 
So,  k = 2.

Answer 2

Hey mate here is ur answer........

x 2 + k (2× + k - 1) + 2 = 0

x 2 + 2kx + k k - 1 + 2 = 0

4k 2 + 4 k - 8

4k - 8 = 4k - 2

For equal roots D is zero

4 ( k - 2 ) = 0

Hence k = 2

Hope it helps♥