Step 1: Set up the system using f(1) and f(2).
Let $ p = f(1) $ and $ q = f(2) $. The two conditions become: $ p + 2q = 0 $ and $ 2p + q = 0 $.
Step 2: Solve the system for p and q.
From the first equation: $ p = -2q $. Substitute into the second: $ 2(-2q) + q = 0 $, so $ -3q = 0 $, giving $ q = 0 $ and then $ p = 0 $. So $ f(1) = 0 $ and $ f(2) = 0 $.
Step 3: Write the conditions for f(x) = ax^2 + bx + c.
From $ f(1) = 0 $: $ a + b + c = 0 $. From $ f(2) = 0 $: $ 4a + 2b + c = 0 $.
Step 4: Subtract the first equation from the second.
$ (4a + 2b + c) - (a + b + c) = 0 - 0 $, giving $ 3a + b = 0 $.
Step 5: Verify this makes sense.
The result $ 3a + b = 0 $ means $ b = -3a $. We can pick any $ a \neq 0 $ and find consistent $ b $ and $ c $. For example, $ a = 1, b = -3, c = 2 $ gives $ f(x) = x^2 - 3x + 2 = (x-1)(x-2) $, and indeed $ f(1) = f(2) = 0 $.
Step 6: State the final answer.
\[ \boxed{3a + b = 0} \]