设方程两根分别为x1,x2
X^2—√2005X—520=0
则由韦达定理得：
X1+X2=√2005,x1x2=-520
由于：A=|x1|+|x2|
则：A^2=(|x1|+|x2|)^2
=|x1|^2+|x2|^2+2|x1||x2|
=x1^2+x2^2+2|x1x2|
=(x1^2+x2^2+2x1x2)-2x1x2+2|x1x2|
=(x1+x2)^2-2x1x2+2|x1x2|
=2005-2(-520)+2|-520|
=2005+1040+1040
=4085