for what values of k are the roots of kx^2+(k-1)x+(k-1) =0 equals
Answers
Answered by
2
For equal roots,
Discriminent = 0
=> b^2 - 4ac = 0
=> (k-1)^2 - 4(k-1)(k) = 0
=> (k-1)^2 = 4(k-1)(k)
=> k - 1 = 4k
=> - 3k = 1
=> k = - 1/3
Discriminent = 0
=> b^2 - 4ac = 0
=> (k-1)^2 - 4(k-1)(k) = 0
=> (k-1)^2 = 4(k-1)(k)
=> k - 1 = 4k
=> - 3k = 1
=> k = - 1/3
snandhini24:
hii
Answered by
0
For equal roots,
Discriminent = 0
=> b^2 - 4ac = 0
=> (k-1)^2 - 4(k-1)(k) = 0
=> (k-1)^2 = 4(k-1)(k)
=> k - 1 = 4k
=> - 3k = 1
=> k = - 1/3
brainliest please ❤️
Similar questions