Root under Root

Observe that

$\sqrt { x+10-6\sqrt { x+1 } } =\sqrt { { \sqrt { x+1 } }^{ 2 }+{ 3 }^{ 2 }-2(\sqrt { x+1 } )3 } =\left| \sqrt { x+1 } -3 \right|$ .

Similarly

$\sqrt { x+2-\sqrt { x+1 } } =\sqrt { { \sqrt { x+1 } }^{ 2 }+{ 1 }^{ 2 }-2(\sqrt { x+1 } )1 } =\left| \sqrt { x+1 } -1 \right|$ .

$\left| \sqrt { x+1 } -3 \right|$ changes its definition at $x=8$ .

Similarly $\left| \sqrt { x+1 } -1 \right|$ changes its definition at $x=0$ .

Now make intervals about $x=0$ and $x=8$ and solve to get $x=0,1,2,3,4,5,6,7,8$ .

By rearrangement:

$\sqrt{(x+1) - 6\sqrt{x+1} +9} + \sqrt{(x+1) - 2\sqrt{x+1} +1}=2$

$\sqrt{(\sqrt{x+1}-3)^2} + \sqrt{(\sqrt{x+1}-1)^2}=2$

$\Rightarrow |\sqrt{x+1}-3| + |\sqrt{x+1}-1|=2$

Since $x \in$ Whole Numbers $\Rightarrow \sqrt{x+1} \geq 1\Rightarrow \sqrt{x+1}-1 \geq 0 \Rightarrow |\sqrt{x+1}-1| = \sqrt{x+1}-1$

$\Rightarrow |\sqrt{x+1}-3| + \sqrt{x+1}-1=2 \Rightarrow |\sqrt{x+1}-3|=3 - \sqrt{x+1}$

$\therefore \sqrt{x+1} \leq 3 \Rightarrow x+1 \leq 9 \Rightarrow x \leq 8$

$\therefore x =$ {0, 1, 2, 3, 4, 5, 6, 7, 8} $\Rightarrow S_x = \dfrac {8*9}{2} = \boxed{36}$