Answer
Solution set: $\{1,5\}$
Work Step by Step
$\sqrt{3x+1}-\sqrt{x-1}=2$
Add $\sqrt{x-1}$ to both sides
$\sqrt{3x+1}=2+\sqrt{x-1}$
Square both sides
$(\sqrt{3x+1})^{2}=(2+\sqrt{x-1})^{2}$...Apply $(a+b)^{2}=a^{2}+2ab+b^{2}$
$3x+1=4+4\sqrt{x-1}+(x-1)$
$2x-2=4\sqrt{x-1}$
Square both sides
$4x^{2}-8x+4=16(x-1)$
$4x^{2}-8x+4=16x-16$
$4x^{2}-24x+20=0$
$x^{2}-6x+5=0$
$(x-1)(x-5)=0$
Testing $x=1$
$\sqrt{3(1)+1}-\sqrt{1-1}=2-0=2$
... $1$ is a solution.
Testing $x=5$
$\sqrt{3(5)+1}-\sqrt{5-1}=4-2=2$
... $5$ is a solution.
Solution set: $\{1,5\}$
