HALP!

$p(x) = x^2 + c_1 x + c_0 \Rightarrow r_1 + r_2 = -c_1\\ f(x) = x^2 - a x + f_0\\ g(x) = x^2 - b x+ g_0\\ h(x) = x^2 - c x + h_0\\ y(x) = f(x) + g(x) + h(x) = 3x^2 -(a+b+c)x + (f_0 +g_0+h_0)$

$y(x) = x^2 - \dfrac 1 3(a+b+c)x + \dfrac 1 3 (f_0 + g_0 + h_0)\\ \text{This is the same form as $p(x)$ listed above with $c_1=-\dfrac 1 3(a+b+c)\\$and thus the sum of the roots is $\dfrac 1 3 (a+b+c)$}$