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Given $|x-2|^{2} +|x-2| - 2=0$

Let $|x-2|=p$

$p^{2}+p-2=0$

$p=1,-2$

Since the roots are real,

$|x-2|=1$

$x=1,3$

Sum of the all real roots of equation$=1+3=4$
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Case 1: If (x-2)>=0, equation will be reduced to

              $(x-2)^{2} + (x-2) - 2 =0$

              i.e. x=0,3

              x=0 will not be accepted as (x-2) will be less than 0

Case 2: If (x-2)<0, equation will be reduced to

              $(-(x-2))^{2} - (x-2) - 2 =0$              

              i.e. x=1,4

             x=4 will not be accepted as (x-2) will be greater than 0

Sum of all real roots = 3+1=4

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