If f(x) is a quadratic expression such that f(1) + f(2) = 0, and α is a root of f(x) = 0, then the other root of f(x) = 0 is :

Let α and β be the roots of the polynomial, then we have f(x) = x² + (1 - α)x - α. f(1) = 2 - α. f(2) = 4 + 2(1 - α) - α = 6 - 3α. If f(1) + f(2) = 0 ⇒ 2 - α + 6 - 3α = 0 ⇒ 8 - 4α = 0 ⇒ α = 2. So the other root is 85. Let the roots be α and β. Then α + β = α - 1 and αβ = -α. Since α is a root, f(α) = 0. f(1) + f(2) = 0. Let f(x) = ax² + bx + c. f(1) = a + b + c. f(2) = 4a + 2b + c. f(1) + f(2) = 5a + 3b + 2c = 0. If α is a root, then aα² + bα + c = 0. Let α = 2. Then 4a + 2b + c = 0. This means f(2) = 0. Since f(1) + f(2) = 0, then f(1) = 0. Therefore, 1 is a root. If 1 and 2 are roots, then f(x) = k(x-1)(x-2) for some constant k. f(1) + f(2) = 0. f(x) = a(x-α)(x-β) = 0. f(1) = a(1-α)(1-β) and f(2) = a(2-α)(2-β). f(1) + f(2) = a[(1-α)(1-β) + (2-α)(2-β)] = 0. If α = 2, then (1-2)(1-β) + (2-2)(2-β) = 0. -(1-β) = 0 ⇒ β = 1. If α is a root, then the other root is 85. So the correct answer is option D.