Absolute value quadratic equation

Expainding square by binomial theoram (Square of a real is always possitive):- $x^{2}-4x+4+|x-2|-2=0$ Now there are two possibilities:- $x^{2}-4x+4-(x-2)-2=0$ And $x^{2}-4x+4+(x-2)-2=0$ Both are quadratic equations and can be easily solved to give following four values of x:- 4,1,0,3 But by the fundamantel theoram of algebra a polynomial of second degree can not have more then two roots.....by subtituting all these four values one by one in the given equation it is easy to see that only roots are 1 and 3 so answer is 1+3=4

replace (x-2 )by v then solve quadratic equation

$\begin{matrix} &\left|x-2\right|^2+\left|x-2\right|-2=0&\\ x-2<0&&x-2\ge 0\\ x<2&&x\ge 2\\\\ \left(-\left(x-2\right)\right)^2-\left(x-2\right)-2=0&&\left(x-2\right)^2+x-2-2=0\\ x^2-5x+4=0&&x^2-3x=0\\ x^2-x-4x+4=0&&x(x-3)=0\\ (x^2-x)+(-4x+4)=0&&x=0,3\\ x(x-1)-4(x-1)=0&&x=3\\ (x-1)(x-4)=0&&\\ x=1,4&&\\ x=1&&\\ &Sum:&\\ &1+3&\\ &=\mathbf{4}&\\ \end{matrix}$

square of mod is always +ve so we'll remove mod from the term which is squared, and then we will take 2 cases for the possible values of Ix-2I 1)in 1st case output is assumed to be +ve and on solvong solutions come as 0,3 but 0 does not satisfy the equation so 3 is a root 2)in the 2nd case output is assumed to be -ve and on solving solutions come as 1,4 but 4 does not satisfy the equation so 1 is a root

from cases (1) and (2) the roots are 1 and 3 and their sum is then 4 So , Ans=>4

consider mod(x-2)=y we get two values of y as 1 and -2 but mod cannot be negative so y=1 so we get two values of x 1&3 so adding we get answer 4