Find the roots of the following quadratic equations by factorisation :

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

Solving,

$\Rightarrow 2x^2 - x + \dfrac{1}{8} = 0 \\[1em] \Rightarrow \dfrac{16x^2 - 8x + 1}{8} = 0 \\[1em] \Rightarrow 16x^2 - 8x + 1 = 0 \\[1em] \Rightarrow 16x^2 - 4x - 4x + 1 = 0 \\[1em] \Rightarrow 4x(4x - 1) - 1(4x - 1) = 0 \\[1em] \Rightarrow (4x - 1)(4x - 1) = 0 \\[1em] \Rightarrow 4x - 1 = 0 \text{ or } 4x - 1 = 0 \\[1em] \Rightarrow 4x = 1 \text{ or } 4x = 1 \\[1em] \Rightarrow x = \dfrac{1}{4} \text{ or } x =\dfrac{1}{4}.$

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