FIND THE VALUE OF K FOR WHICH THE ROOTS OF THE EQUATION (K+4) x^{2} + (K+1) x + 1 = 0 are real and equal.

since for the real and equal roots ,

b^{2} -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

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