Question

Find the value of k in which (k+4)x^2+(k+1)x+1=0 has equal roots

Answer 1

hope you have got the answer
value of k is 5

Answer 2

Here we have

(k + 4)x^2 + (k + 1)x + 1 =0

★For equal roots

b^2 - 4ac = 0

b^2 = 4ac

(k + 1)^2 = 4(k + 4)

k^2 + 2k + 1 - 4k - 16 = 0

k^2 - 2k - 15 = 0

(k - 5)(k + 3) = 0

k = 5

k = -3