Question

find the value of k for which the quadratic equation (k+4)x^2+(k+1)x+1=0 has equal roots

Answer 1

for equal roots b^2-4ac =0
(k+1)^2 - 4(k+4)(1) = 0
k^2 + 1 + 2K - 4K -16 = 0
k^2 -2k -15 = 0
k^2 +3k -5k -15 =0
k (k+3) -5(k+3) =0
(k-5)(k+3)=0
so k=5 , k= -3
hope it helps.