Aus Online Mathematik Brückenkurs 1

Square both sides of the equation so that the root sign disappears,

\displaystyle 1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2

and then solve the resulting second-order equation by completing the square,

\displaystyle \begin{align} x^{2}-3x+3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.} \end{align}

As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how x is chosen) so, the original root equation does not have any solutions.