(i) (k – 12)x2 + 2 (k – 12)x + 2 = 0
Sol. (k – 12)x2 + 2 (k – 12)x + 2 = 0
Comparing with ax2 + bx + c = 0 we have a = k – 12, b = 2 (k – 12), c = 2
We know that,
∆ = b2 – 4ac
= [2 (k – 12)]2 – 4 (k -12) (2)
= (2k – 24)2 – 8 (k – 12)
= 4k2 – 96k + 576 – 8k + 96
= 4k2 – 104k + 672
∵ The roots of given equation are real and equal.
∴ ∆ must be zero.
∴4k2 – 104k + 672 = 0
∴4 (k2 – 26k + 168) = 0
∴ k2 – 14k – 12k + 168 = 0
∴ k (k – 14) – 12 (k – 14)= 0
∴ (k – 14) (k – 12) = 0
∴k – 14 = 0 or k – 12 = 0
∴ k = 14 or k = 12
