The sum of all the real roots of the equation $|\mathrm{x}-2{|}^2+|\mathrm{x}-2|-2=0$ is

Case 1 : Suppose x $\ge$ 2 . Then the equation becomes
$(\mathrm{x}-2{)}^2+(\mathrm{x}-2)-2=0;{\mathrm{x}}^2-3\mathrm{x}=0$
Since x $\ge$ 2 , x = 3
Case 2 : Suppose x < 2. Then the equation becomes
$(2-\mathrm{x}{)}^2+(2-\mathrm{x})-2=0;(\mathrm{x}-1\left)\right(\mathrm{x}-4)=0 \mathrm{x}=1.$
Thus the real roots of the equation are 1 and 3 and their sum is 4.