FIND THE VALUE OF K FOR WHICH THE ROOTS OF THE EQUATION (K+4) x2 + (K+1) x + 1 = 0 are real and equal.
since for the real and equal roots ,
b2 -4ac =0
(k+1)2 - 4 x (K+4) x 1 =0
k2 +2k +1 -4k - 16 =0
k2 -2k -15 = 0
k2 -5k + 3k -15 =0
k(k-5) + 3 (k-5) =0
(k-5) (k+3) =0
then , k = 5 and k =-3
HOPE THIS HELPS...