Алгебра: корень(3x+1)-корень(x-1)=2

ОДЗ: $\left \{ {{3x+1 \geq 0} \atop {x-1 \geq 0}} \right. \ \ \left \{ {{x \geq - \frac{1}{3} } \atop {x \geq 1}} \right. \ \ \Rightarrow x \geq 1$

$\sqrt{3x+1} - \sqrt{x-1} = 2 \\ \\ \sqrt{3x+1} = 2+ \sqrt{x-1} \\ \\ ( \sqrt{3x+1})^2 = (2+ \sqrt{x-1})^2 \\ \\ 3x+1 = 4 + 2\sqrt{x-1} + x-1 \\ \\ x-1 = 2\sqrt{x-1} \\ \\ (x-1)^2 = (2\sqrt{x-1} )^2 \\ \\ x^2 - 2x+1 = 4x-4 \\ \\ x^2-6x+5 = 0$
Корни уравнения
$x_{1} = 5 ; \ x_{2} = 1$

Ответ: $x_{1} = 5 ; \ x_{2} = 1$