10.1. Find all values of the parameter $a$ for which the equation $\left|x^{2}-2 a x+1\right|=a$ has three roots.
Answer: $a=\frac{1+\sqrt{5}}{2}$. Solution. Obviously, $a \geq 0$ (since the left side of the equation is non-negative). If the discriminant of the quadratic trinomial $D=4\left(a^{2}-1\right)$ is negative, then we have the quadratic equation $x^{2}-2 a x+1=a$, and, therefore, it has no more than two roots. Therefore, $D \geq 0$, i.e., $a \geq 1$, and the equation will have three roots when the line $y=a$ is tangent to the graph of $y=\left|x^{2}-2 a x+1\right|$ at the "reflected" vertex of the parabola (i.e., is tangent to the graph of $y=-\left(x^{2}-2 a x+1\right)$, obtained after symmetric reflection relative to the $O x$ axis). The vertex of the parabola has an abscissa $x=a$, so $a=-\left(a^{2}-2 a^{2}+1\right) \Leftrightarrow a=\frac{1+\sqrt{5}}{2}$ (since $\left.a \geq 1\right)$.