Correct Answer – 4
Given, `|x-2|^(2)+|x-2|-2=0`
Case I When `c ge 2`
`implies(x-2)^(2)+(x-2)-2=0`
`impliesx^(2)+4-4x+x-2-2=0`
`impliesx^(2)-3x=0`
`impliesx(x-3)=0`
`impliesx=0,3″ “[0 “is rejected”]`
`impliesx=3″ “…(i)`
Case II When `x gt 2`
`implies{-(x-2))^(2)-(x-2)-3=0`
`(x-2)^(2)-x+2-2=0`
`impliesx^(2)+4-4x-x=0`
`impliesx^(2)–4x-(x-4)=0`
`impliesx(x-4)-1(x-4)=0`
`implies(x-1)(x-4)=0`
`impliesx=1,4″ “[“4 is rejected”]`
`impliesx=1″ “…(ii)`
Hence, the sum of the roots is `3+1 =4.`
Alternate Solution
Given, `|x-2|^(2)+|x2|2=0`
`implies(|x-2|+2)(|x-2|-1)=0`
`therefore|x-2|=-2,1″ “[“neglacting”-2]`
`implies|x-2|=1impliesx=3,1`
`implies` Sum of the roots `=4`