(c) $2x^2 - \dfrac{4}{2}x + 1 = 0$
For $ax^2 + bx + c = 0$, sum of roots $= -\dfrac{b}{a}$. Here, $-\dfrac{b}{a} = -\dfrac{(-2)}{2} = \dfrac{2}{2}$...
Rewriting: $2x^2 - 2x + 1 = 0$, so sum $= -\dfrac{-2}{2} = 1$.
Re-checking (a): $2x^2 - 4x + 8 = 0 \Rightarrow$ sum $= -\dfrac{-4}{2} = \mathbf{4}$. ✓
Answer: (a) $2x^2 - 4x + 8 = 0$
For a quadratic $ax^2 + bx + c = 0$, sum of roots $= -b/a$.
Always divide $-b$ by $a$ (not just read off the coefficient). Option (a) is the correct answer.