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

(i) kx(x - 2) + 6 = 0

(ii) (k + 4)x² + (k + 1)x + 1 = 0

(i) kx(x - 2) + 6 = 0

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 = $\dfrac{6}{6}$ or x = 1

⇒ x = 1 equation has real and equal roots

∴ k = 6

Hence, k = 6.

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Given,

(k + 4)x² + (k + 1)x + 1 = 0

Comparing (k + 4)x² + (k + 1)x + 1 = 0 with ax² + bx + c = 0 we get,

a = k + 4, b = k + 1 and c = 1.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (k + 1)² - 4(k + 4)(1) = 0

⇒ k² + 2k + 1 - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k + 3)(k - 5) = 0

⇒ k + 3 = 0 or k - 5 = 0

⇒ k = -3 or k = 5.

If k = 5, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (5 + 4)x² + (5 + 1)x + 1 = 0

⇒ 9x² + 6x + 1 = 0

⇒ (3x + 1)² = 0

⇒ 3x + 1 = 0

⇒ x = $\dfrac{-1}{3}$

Both roots are the same.

If k = -3, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (-3 + 4)x² + (-3 + 1)x + 1 = 0

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

⇒ x - 1 = 0

⇒ x = 1

Both roots are the same.

Hence, k = -3 or 5.