k2x2 – 2 (k – 1)x + 4 = 0

(ii) k²x² – 2 (k – 1)x + 4 = 0

Sol. k²x² – 2 (k – 1)x + 4 = 0

Comparing with ax² + bx + c = 0 we have a = k², b = – 2 (k – 1), c = 4

We know that,

∆  = b² – 4ac

= [– 2 (k – 1)]² – 4 (k²) (4)

= (– 2k + 2)² – 16k²

= 4k² – 8k + 4 – 16k²

= – 12k² – 8k + 4

∵  The roots of given equation are real and equal.

∴ ∆  must be zero.

∴  – 12k² – 8k + 4 = 0

∴  – 4 (3k² + 2k – 1) = 0

∴   3k² + 3k – k – 1 =  -⁰/₄

∴  3k (k + 1) – 1 (k + 1) = 0

∴  (k + 1) (3k – 1) = 0

∴  k + 1 = 0 or 3k – 1 = 0

∴  k = – 1 or 3k = 1

∴ k = -1 or k = ¹/₃