# Task 5. (20 points)

Find all values of the parameter $a$ for which the roots $x_{1}$ and $x_{2}$ of the equation

$$
2 x^{2}-2016(x-2016+a)-1=a^{2}
$$

satisfy the double inequality $x_{1}<a<x_{2}$.

#
# Solution.

Let the polynomial from the problem statement be denoted by $f(x)$. Then, due to the positivity of the coefficient of $x^{2}$, the condition of the problem is equivalent to the simple inequality $f(a)<0$. That is, $f(a)=a^{2}-2a \cdot 2016+2016^{2}-1=(a-2016)^{2}-1<0$. From this, we obtain that $a \in (2015,2017)$.