(v) (k - 12) x2 + 2 (k - 12) x ·+ 2 = 0 . ( vz) x2 - 4kx + k = 0

1.	(v) (k - 12) x2 + 2 (k - 12) x ·+ 2 = 0 . ( vz) x2 - 4kx + k = 0
Answer» Here, a = k – 12, b = 2 (k – 12), c = 2∴ D = b2\xa0– 4ac = [2(k– 12)]2– 4(k – 12) x 2= 4 (k – 12)2\xa0– 8 (k– 12)Roots are equal, if D = 0⇒ 4 (k – 12) 2 – 8 (k – 12) = 0⇒ 4(k – 12)(k – 12 – 2) = 0⇒ (k – 12) (k – 14) = 0⇒ k = 12 or 14But k = 12 does not satisfy the eqn.k = 14\xa0Ans.

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