(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
