For what values of k, the roots of the quadratic equation (k+4)x²+(k+1)x+1 = 0 are equal?

The given Quadratic Equation is-

(k+4)x²+(k+1)x+1 = 0

For Equal Roots, D should be equal to zero.

Discriminant D of the quadratic equation ax²+bx+c = 0 is given by-

D = b²-4ac

Comparing the equation ax²+bx+c = 0 with given quadratic equation (k+4)x²+(k+1)x+1 = 0, we get-

a = k+4 , b = k+1 and, c = 1

∴ D = 0

⇒ (k+1)²-4(k+4) = 0

⇒ k²+2k+1-4k-16 = 0

⇒ k²-2k-15 = 0

⇒ k²-5k+3k-15 = 0

⇒ k(k-5)+3(k-5) = 0

⇒ (k+3)(k-5) = 0

∴ k=-3 or k=5

Hence, the possible values of k are -3 and 5.