Answer
Both pieces pieces are $4 \ in$ long.
Work Step by Step
$(\dfrac{x}{4})(\dfrac{x}{4}) + (\dfrac{8-x}{4})(\dfrac{8-x}{4}) = 2 \ in \longrightarrow \dfrac{x^2}{16} + \dfrac{(8-x)^2}{16} = 2 \longrightarrow \dfrac{x^2 + 8^2 - 2(8)(x) +x^2}{16} = 2 \longrightarrow \dfrac{2x^2 - 16x +64}{16} = 2 \longrightarrow \dfrac{x^2-8x+32}{8} = 2 \longrightarrow x^2-8x+32=16 \longrightarrow x^2-8x+32-16=0 \longrightarrow x^2-8x+16=0 \longrightarrow (x-4)^2=0 \longrightarrow \sqrt{(x-4)^2} = \sqrt{0} \longrightarrow x-4=0 \longrightarrow x = 4 \ in$
Since $x = 4 $ both pieces pieces are $4 \ in$ long.
