1. Solve $|x-2|^2 + |x-2| - 2 = 0$

Let $y=|x-2|$ Then $y\ge 0$, and the equation becomes $y^2+y-2=0$ Factorizing, $(y+2)(y-1)=0$ So, $y=1 \quad \text{or} \quad y=-2$ Since $y\ge 0$, only $y=1$ is valid.

Thus, $|x-2|=1$ So, $x-2=\pm 1$ Hence roots are $x=3,\;1$ Their squares are $3^2=9,\quad 1^2=1$ Sum of squares from first equation: $9+1=10$

2. Solve $x^2-2|x-3|-5=0$

We split into cases.

Case 1: $x\ge 3$

Then $|x-3|=x-3$. Equation becomes $x^2-2(x-3)-5=0$ $x^2-2x+6-5=0$ $x^2-2x+1=0$ $(x-1)^2=0$ So $x=1$. But this does not satisfy $x\ge 3$, so reject.

Case 2: $x<3$

Then $|x-3|=3-x$. Equation becomes $x^2-2(3-x)-5=0$ $x^2-6+2x-5=0$ $x^2+2x-11=0$ So, $x=\frac{-2\pm\sqrt{4+44}}{2}=\frac{-2\pm\sqrt{48}}{2}=\frac{-2\pm 4\sqrt{3}}{2}=-1\pm 2\sqrt{3}$ Now check $x<3$:

• $-1+2\sqrt{3}\approx 2.464<3$ valid
• $-1-2\sqrt{3}<3$ valid

So roots are $x=-1+2\sqrt{3},\quad -1-2\sqrt{3}$

Their squares sum is $(-1+2\sqrt{3})^2+(-1-2\sqrt{3})^2$ Use identity: $(a+b)^2+(a-b)^2=2(a^2+b^2)$ Here $a=-1$, $b=2\sqrt{3}$. So, $2\left[(-1)^2+(2\sqrt{3})^2\right]=2(1+12)=26$

3. Total sum

$10+26=36$

So the correct option is $\boxed{\text{C: }36}$

4. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C.