Find the value of k for which the roots of the following equation are real and equal.

k²x² - 2(2k - 1)x + 4 = 0

Given,

Equation : k²x² - 2(2k - 1)x + 4 = 0

The roots of the equation are real and equal, when D = 0.

⇒ b² - 4ac = 0

⇒ [-2(2k - 1)]² - 4 × k² × 4 = 0

⇒ [-4k + 2]² - 16k² = 0

⇒ 16k² + 4 - 16k - 16k² = 0

⇒ 16k = 4

⇒ k = $\dfrac{4}{16} = \dfrac{1}{4}$.

Hence, k = $\dfrac{1}{4}$.