Question

Determina numerele reale x si y pt. care x² 6√2 x+y² +4y +22=0.

Answer 1

22 = 4 +18

$\it \Leftrightarrow (y^2+4y+4) + (x^2-6\sqrt2x+18)=0$

$\it \Leftrightarrow (y+2)+(x-3\sqrt2)^2=0$

$\it y+2=0 \Longrightarrow y = -2$

$\it x-3\sqrt2 = 0 \Longrightarrow x =3\sqrt2.$