[ Quadratic equations. Root formula ]  $[$ Symmetry helps solve the task. $]$

Given a reduced quadratic trinomial $f(x)=x^{2}+b x+c$, having two distinct roots. Denote by $D$ its discriminant ( $D=b^{2}-4 c$ ). How many roots does the equation $f(x)+f(x-\sqrt{D})=0$ have?
How are the graphs of the functions $f(x)$ and $f(x-\sqrt{D})$ related?

## Solution

Let $d=\sqrt{D}, x_{1}, x_{2}\left(x_{1}<x_{2}\right)$ be the roots of the equation $f(x)=0$. From the formula for the roots, it is clear that $x_{2}=x_{1}+d$. The graph of the function $f(x-d)$ is obtained from the graph of the function $f(x)$ by shifting it by $d$ to the right along the $O x$ axis. Therefore, first, $x_{2}$ is a root of the equation $f(x-d)=0$ (and thus a root of the equation

$f(x)+f(x-d)=0)$, and secondly, the graphs of the functions $f(x)$ and $f(x-d)$ are symmetric with respect to the vertical line $x=x_{2}$, and therefore, the graph of the function

$f(x)+f(x-d)$ is also symmetric with respect to this line. We obtain that the quadratic trinomial $f(x)+f(x-d)$ has an odd number of roots (one of the roots is $x_{2}$, and the others are paired symmetrically with respect to $x_{2}$). Since a quadratic equation has no more than two roots, the only possibility is that the equation $f(x)+f(x-d)=0$ has exactly one root.

## Answer

One root.