解：?$x + \sqrt {x - 1} = 3，$?移項得?$\sqrt {x - 1} = 3 - x$?
?$ $?兩邊平方得?$x - 1 = (3 - x)^2$?
整理得?$x^2 - 7x + 10 = 0$?
?$ $?因式分解得?$(x - 2)(x - 5) = 0$?
?$ $?解得?$x_{1} = 2，$??$x_{2} = 5$?
檢驗：當?$x = 2$?時，左邊?$= 2 + \sqrt {2 - 1} = 3，$?右邊?$= 3，$?
左邊?$=$?右邊，?$x = 2$?是原方程的根
?$ $?當?$x = 5$?時，左邊?$= 5 + \sqrt {5 - 1} = 7\neq 3，$??$x = 5$?是增根，舍去
∴原方程的根是?$x = 2$?
