(k – 12)x2 + 2 (k – 12)x + 2 = 0

(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