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

(k + 1)x² + 2(k + 3)x + (k + 8) = 0

Comparing (k + 1)x² + 2(k + 3)x + (k + 8) = 0 with ax² + bx + c = 0 we get,

a = (k + 1), b = 2(k + 3) and c = (k + 8).

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 3)]² - 4 × (k + 1) × (k + 8) = 0

⇒ 4(k + 3)² - (4k + 4) × (k + 8) = 0

⇒ 4[(k)² + (3)² + 2 × k × 3] - (4k² + 32k + 4k + 32) = 0

⇒ 4(k² + 9 + 6k) - (4k² + 36k + 32) = 0

⇒ 4k² + 36 + 24k - 4k² - 36k - 32 = 0

⇒ -12k + 4 = 0

⇒ -12k = -4

⇒ k = $\dfrac{-4}{-12}$

⇒ k = $\dfrac{1}{3}$.

Hence, k = $\Big{\dfrac{1}{3}\Big}$.