Question

From the following equation, find the value of constant 'k' so that the equation has real and equal roots.

kx(x − 2) + 6 = 0

Answer 1

Given,

kx(x − 2) + 6 = 0

kx² − 2kx + 6 = 0

Comparing kx² − 2kx + 6 = 0 with ax² + bx + c = 0 we get,

a = k, b = −2k and c = 6.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² − 4ac = 0

⇒ (-2k)² − 4(k)(6) = 0

⇒ 4k² − 24k = 0

⇒ 4k(k − 6) = 0

⇒ 4k = 0 or (k − 6) = 0

⇒ k = 0 or k = 6

If k = 0, substituting in L.H.S. of kx² − 2kx + 6 = 0 we get,

= (0)x² − 2(0)x + 6

= 6

L.H.S. ≠ R.H.S.

Thus, k ≠ 0.

Substituting k = 0 in kx² − 2kx + 6 = 0 we get,

⇒ (6)x² − 2(6)x + 6 = 0

⇒ 6x² − 12x + 6 = 0

⇒ 6x² − 6x − 6x + 6 = 0

⇒ 6x(x − 1) − 6(x − 1) = 0

⇒ (6x − 6)(x − 1) = 0

⇒ (6x − 6) = 0 or (x − 1) = 0

⇒ x = `6/6`​ or x = 1

⇒ x = 1 equation has real and equal roots

∴ k = 6