Question

If the roots of the given quadratic equation are real and equal, then find the value of ‘k’

kx(x – 2) + 6 = 0

Answer 1

kx(x – 2) + 6 = 0

∴ kx² – 2kx + 6 = 0

Comparing the above equation with

ax² + bx + c = 0, we get

a = k, b = – 2k, c = 6

∆ = b² – 4ac

= (–2k)² – 4 × k × 6

= 4k² – 24k

∴ ∆ = 4k(k – 6)

Since the roots are real and equal,

∆ = 0

∴ 4k(k – 6) = 0

∴ k(k – 6) = 0

∴ k = 0 or k – 6 = 0

But, if k = 0, then quadratic coefficient becomes zero.

∴ k ≠ 0

∴ k = 6