The roots of the quadratic equation $2x^2 - x + \dfrac{1}{8} = 0$ are

The correct answer is $\dfrac{1}{4}, \dfrac{1}{4}$

$2x^2 - x + \dfrac{1}{8} = 0$

⇒ $16x^2 - 8x + 1 = 0$

⇒ $16x^2 - 4x - 4x + 1 = 0$

⇒ $16x(x - \dfrac{1}{4}) -4(x - \dfrac{1}{4}) = 0$

⇒ $(16x - 4)(x - \dfrac{1}{4}) = 0$

⇒ $x = \dfrac{1}{4}, \dfrac{1}{4}$