Question

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

kx² + (6k + 2)x + 16 = 0

Answer 1

kx² + (6k + 2)x + 16 = 0

a = k, b = (6k + 2), c = 16

Δ = b² – 4ac   ....[∵ the roots are real and equal]

⇒ (6k + 2)² – 4 × k × 16 = 0

⇒ 36k² + 24k + 4 – 64k = 0

⇒ 36k² – 40k + 4 =0

⇒ 36k² – 36k – 4k + 4 = 0

⇒ 36k (k – 1) – 4 (k – 1) = 0

⇒ 4 (k – 1) (9k – 1) = 0

⇒ k = 1 or k = `1/9`