Find that value of k in the following quadratic equation whose roots are real and equal:

x² – 2(k + 1)x + k² = 0

Let us first solve the above equation,

x² – 2kx + 1x + k² = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 1

b = -2(k+1)

c = k²

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (-2(k+1)) ² – (4 × k² × 1) = 0

⟹ 4(k² + 2k + 1) – 4 k² = 0

⟹ 4k² + 8k + 4 – 4 k² = 0

⟹ 8k + 4 = 0

⟹ 8k = - 4

⟹ k = - 4/8

⟹ k = - 1/2