Condition : \(x\ne0\)

\(x+\dfrac{2a\left|x+a\right|}{x}=\dfrac{a^2}{x}\)\(\Leftrightarrow\dfrac{x^2+2a\left|x+a\right|-a^2}{x}=0\)

\(\Leftrightarrow x^2+2a\left|x+a\right|-a^2=0\)

If \(x+a\ge0\Leftrightarrow x\ge-a\), then :

\(x^2+2a\left(x+a\right)-a^2=0\Leftrightarrow x^2+2ax+2a^2-a^2=0\)

\(\Leftrightarrow x^2+2ax+a^2=0\Leftrightarrow\left(x+a\right)^2=0\Leftrightarrow x=-a\)

If \(x+a< 0\Leftrightarrow x< -a\), then :

\(x^2-2a\left(x+a\right)-a^2=0\Leftrightarrow x^2-2ax-2a^2-a^2=0\)

\(\Leftrightarrow x^2-2ax+a^2-4a^2=0\Leftrightarrow\left(x-a\right)^2-\left(2a\right)^2=0\)

\(\Leftrightarrow\left(x-3a\right)\left(x+a\right)=0\)

Since \(x+a< 0,x-3a=0\) or x = 3a. Moreover, x + a < 0. So, a < 0 

Hence, x = -a or x = 3a (only when a < 0)