The solution set for the quadratic equation 2x² - x + $\dfrac{1}{8}$ = 0 is:

1. $\Big{\dfrac{1}{4}, \dfrac{1}{4}\Big}$

2. $\Big{-\dfrac{1}{4}, \dfrac{1}{4}\Big}$

3. $\Big{-\dfrac{1}{2}, \dfrac{1}{4}\Big}$

4. {4, 4}

Given,

⇒ 2x² - x + $\dfrac{1}{8}$ = 0

Multiply the equation with 16, we get:

⇒ 16(2x² - x + $\dfrac{1}{8}$ = 0)

⇒ 32x² - 16x + $\dfrac{16}{8}$ = 0

⇒ 32x² - 16x + 2 = 0

⇒ 2(16x² - 8x + 1) = 0

⇒ 16x² - 8x + 1 = 0

⇒ 16x² - 4x - 4x + 1 = 0

⇒ 4x(4x - 1) - 1(4x - 1) = 0

⇒ (4x - 1)(4x - 1) = 0

⇒ (4x - 1) = 0 or (4x - 1) = 0      [Using Zero-product rule]

⇒ 4x = 1 or 4x = 1

⇒ x = $\dfrac{1}{4}$ or x = $\dfrac{1}{4}$

Hence, option 1 is the correct option.