A problem by Δημήτριος Χαλάτσης

From the gi \ven equality about (α-2)( $\overline{α}$ -2)+|α-2|=2 <=> |α-2|=1 => |α|<3 then $α_{2}u^{2}$ + $α_{1}u$ + $α_{0}$ =0 <=> $u^{3}$ + $α_{2}u^{2}$ + $α_{1}u$ + $α_{0}$ =- $u^{3}$
=> | $α_{2}u^{2}$ + $α_{1}u$ + $α_{0}$ |=| $u^{3}$ | through the triangular inequality we have that

| $α_{2}u^{2}$ |+| $α_{1}u$ |+| $α_{0}$ | > | $u^{3}$ | since |α|< 3

| $3u^{2}$ | + | $3u$ | + 3 > | $α_{2}u^{2}$ |+| $α_{1}u$ |+| $α_{0}$ | > | $u^{3}$ | =>

| $3u^{2}$ | + | $3u$ | + 3 > | $u^{3}$ | <=> | $u^{3}$ | - | $3u^{2}$ | - | $3u$ | - 3 < 0 using horners method to factorize the "polynomial" with | $u$ | - 4 (| $u$ | - 4)( | $u^{2}$ | + | $u$ | + 1) + 1 < 0 <=> (| $u$ | - 4)( | $u^{2}$ | + | $u$ | + 1) < -1 => (| $u$ | - 4)( | $u^{2}$ | + | $u$ | + 1) < 0 since ( | $u^{2}$ | + | $u$ | + 1) > 0 always | $u$ | - 4 < 0 <=> | $u$ | < $4$