Question

Find the value of ‘k’ to identify the roots of the following equation is real and equal

(5k – 6)x² + 2kx + 1 = 0

Answer 1

(5k – 6)x² + 2kx + 1 = 0

a = (5k – 6), b = 2k, c = 1

Δ = b² – 4ac

⇒ (2k)² – 4(5k – 6)(1)

⇒ 4k² – 20k + 24 = 0  ...[∵ Since the roots are real and equal]

⇒ k² – 5k + 6 = 0

⇒ (k – 3)(k – 2) = 0

k = 3, 2