Answer
$x=8$
Work Step by Step
$\sqrt{3x+1}=2+\sqrt{x+1}$ (Take the square)
$3x+1=4+(x+1)+4\sqrt{x+1}$ (Simplify)
$3x+1=x+5+4\sqrt{x+1}$
$2x-4=4\sqrt{x+1}$ (Divide by 2)
$x-2=2\sqrt{x+1}$ (Take the square again)
$x^2-4x+4=4(x+1)$
$x^2-4x+4=4x+4$
$x^2-8x=0$
$x(x-8)=0$
$x=0$ or $x=8$
Let us verify by back substituting to the original equation.
For $x=0$,
$\sqrt{3\cdot 0+1}=2+\sqrt{0+1}$
$\sqrt{1}=2+\sqrt{1}$
$1=3$ (False)
For $x=8$,
$\sqrt{3\cdot 8+1}=2+\sqrt{8+1}$
$\sqrt{25}=2+\sqrt{9}$
$5=5$ (True)
So, the real solution is only $x=8$.
