[ Quadratic equations. Vieta's theorem ]

Given quadratic trinomials $f$ and $g$ with the same leading coefficients. It is known that the sum of the four roots of these trinomials

is equal to $p$. Find the sum of the roots of the trinomial $f+g$, if it is known that it has two roots.
Let $f(x)=a x^{2}+b_{1} x+c_{1}, g(x)=a x^{2}+b_{2} x+c_{2}$. By the condition, $p=-b_{1 / a}-b_{2 /}$. Since $f(x)+g(x)=2 a x^{2}+\left(b_{1}+\right.$ $\left.b_{2}\right) x+\left(c_{1}+c_{2}\right)$, the sum of the roots of this quadratic polynomial is $-b_{1}+b_{2 / 2}=p / 2$.

Answer

## $p / 2$.