If the roots of the quadratic equation (k + 1)x² — 2(k—1)x + 1 = 0 are real and equal, find the value of k.

Comparing equation (k + 1)x² — 2(k—1)x + 1 = 0 with ax² + bx + c = 0 we get

a = (k + 1), b = – 2(k – 1) and c = 1

Discriminant (D) = b² – 4ac

As roots are real and equal D = 0

⇒ b² – 4ac = 0

⇒ [ – 2(k – 1)]² – 4(k + 1)(1) = 0

⇒ (4)(k – 1)² – 4k – 4 = 0

Expand (k – 1)² using (a – b)² = a² – 2ab + b²

⇒ 4(k² – 2k + 1) – 4k – 4 = 0

⇒ 4k² – 8k – 4k = 0

⇒ 4k² – 12k = 0

Take 4k factor common

⇒ 4k(k – 3) = 0

⇒ 4k = 0 or k – 3 = 0

⇒ k = 0 or k = 3

Therefore k = 0 and k = 3