10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2016$ and $a \neq c$ (the numbers themselves are not given).
Solution. Expanding the brackets $\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=\mathrm{x}^{2}-(\mathrm{c}+\mathrm{d}) \mathrm{x}+\mathrm{cd}$, i.e., $\quad(c+d-a-b) x=c d-a b$. By the condition $c - a = d - b$ - let's denote this by $r$. Then $a=c-r, d=b+r$ and $c d-a b=c(b+r)-(c-r) b=(b+c) r=2016 r$. Thus (since $c - -\mathrm{r}, \mathrm{d}-\mathrm{b}=\mathrm{r}) \quad 2 \mathrm{rx}=2016 \mathrm{r}, \quad \mathrm{x}=1008 \quad(\mathrm{r} \neq 0$ by the condition).

Remark. There are other solutions. For example: Let $f(x)=(x-a)(x-b)$, $g(x)=(x-c)(x-d)$. Then the condition $a+d=b+c$ gives the identity $g(x)=f(x+a-c)$, and the equation $f(x)=g(x)$ reduces to the equation $f(x)=f(x+a-c)$, which means the symmetry of the points $x$ and $x+a-c$ with respect to the abscissa $x_{0}=\frac{a+b}{2}$ of the vertex of the parabola $y=f(x)$, from which we obtain

$$
\frac{x+x+a-c}{2}=\frac{a+b}{2}, \quad \mathrm{x}=\frac{b+c}{2}=\frac{2016}{2}=1008
$$

Answer: 1008.