If x2 + k (4x + k – 1) + 2 = 0 has equal roots, then k =A. \(-\frac

1.	If x2 + k (4x + k – 1) + 2 = 0 has equal roots, then k =A. \(-\frac{2}{3},1\)B. \(\frac{2}{3},-1\)C. \(\frac{3}{2},\frac{1}{3}\)D. \(-\frac{3}{2},-\frac{1}{3}\)
Answer» Equation x² + k (4x + k – 1) + 2 = 0 has equal roots d = 0 d = b² – 4ac = 0 Here a = 1, b = 4k, c = k² – k + 2 ⇒ 16 k² – 4(k² – k + 2) = 0 ⇒ 12 k² + 4k – 8 = 0 ⇒ 3k2 + k – 2 = 0 ⇒ (3k – 2) (k + 1) = 0 ⇒ K = 2/3, – 1

If x2 + k (4x + k – 1) + 2 = 0 has equal roots, then k =A. \(-\frac{2}{3},1\)B. \(\frac{2}{3},-1\)C. \(\frac{3}{2},\frac{1}{3}\)D. \(-\frac{3}{2},-\frac{1}{3}\)

Answer»

Equation x² + k (4x + k – 1) + 2 = 0 has equal roots

d = 0

d = b² – 4ac = 0

Here a = 1, b = 4k, c = k² – k + 2

⇒ 16 k² – 4(k² – k + 2) = 0

⇒ 12 k² + 4k – 8 = 0

⇒ 3k2 + k – 2 = 0

⇒ (3k – 2) (k + 1) = 0

⇒ K = 2/3, – 1