For the above expression to be a perfect square, D = b² – 4ac = 0

⇒ (2k + 4)² – 4 × (4 – k)(8k + 1) = 0

⇒ 4k² + 16k + 16 + 32k² – 124k – 16 = 0

⇒ 36k² – 108k = 0

⇒ 36k(k – 3) = 0

⇒ k = 0, 3