Question

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

Answer 1

Answer:

k=2

Step-by-step explanation:

x² + K (2x + K-1) +2=0

x²+2kx+(k²-k+2)=0

here a=1,b=2k and c=k²-k+2

For roots to be equal

b²=4ac

(2k)²=4*1(k²-k+2)

4k²=4K²-4k+8

4k=8

k=8/4=2