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