Question

Find tje 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:

x^2 + k(2x + k-1) + 2 = 0

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

to have equal roots

(2k)^2 = 4 (k^2-k+2)

4k^2 = 4k^2-4k+8

4k = 8

k = 2