Find the values of k for which the following equation has equal roots:

(3k + 1)x² + 2(k + 1)x + k = 0

Comparing (3k + 1)x² + 2(k + 1)x + k = 0 with ax² + bx + c = 0 we get,

a = (3k + 1), b = 2(k + 1) and c = k.

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 1)]² - 4 × (3k + 1) × (k) = 0

⇒ 4(k + 1)² - (12k + 4) × (k) = 0

⇒ 4[(k)² + (1)² + 2 × k × 1] - (12k² + 4k) = 0

⇒ 4(k² + 1 + 2k) - 12k² - 4k = 0

⇒ 4k² + 4 + 8k - 12k² - 4k = 0

⇒ -8k² + 4k + 4 = 0

⇒ -8k² + 8k - 4k + 4 = 0

⇒ -8k(k - 1) - 4(k - 1) = 0

⇒ (k - 1)(-8k - 4) = 0

⇒ (k - 1) = 0 or (-8k - 4) = 0      [Using Zero-product rule]

⇒ k = 1 or -8k = 4

⇒ k = 1 or k = $\dfrac{4}{-8}$

⇒ k = 1 or k = $-\dfrac{1}{2}$

Hence, k = $\Big{-\dfrac{1}{2}, 1\Big}$.