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

kx² + kx + 1 = -4x² - x

⇒ kx² + kx + 1 = -4x² - x

⇒ kx² + kx + 1 + 4x² + x = 0

⇒ kx² + 4x² + kx + x + 1 = 0

⇒ x² (k + 4) + x(k + 1) + 1 = 0

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

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

Since equations has equal roots,

∴ D = 0

⇒ (k + 1)² - 4.(k + 4).1 = 0

⇒ [(k)² + (1)² + 2 × k × 1] - (4k + 16) = 0

⇒ k² + 1 + 2k - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k - 5)(k + 3) = 0

⇒ (k - 5) = 0 or (k + 3) = 0      [Using Zero-product rule]

⇒ k = 5 or k = -3

Hence, k = {5 , -3}.