1. It is known that the quadratic function $f(x)=x^{2}+a x+b$ has zeros $x_{1}$, $x_{2}$ and $f(2014)=f(2016)$. Find $\frac{x_{1}+x_{2}}{2}$.
# Answer: 2015

## First Solution.

Since the graph of the quadratic function $f(x)$ is symmetric with respect to the line $x=-a / 2$ and $f(2014)=f(2016), f\left(x_{1}\right)=f\left(x_{2}\right)=0$, then

$-\frac{a}{2}=\frac{x_{1}+x_{2}}{2}=\frac{2014+2016}{2}=2015$.

## Second Solution.

Since $f(2014)=f(2016)$, then $2014^{2}+2014 a+b=2016^{2}+2016 a+b \Leftrightarrow$ $2 a=-2 \cdot 4030 \Leftrightarrow a=-4030$. By Vieta's theorem $x_{1}+x_{2}=-a \Rightarrow$ $\frac{X_{1}+X_{2}}{2}=-\frac{a}{2}=2015$

Recommendations for Checking

Only the answer is given: 0 points.

Only the value of $a$ is found: 4 points.