Question

determinati nr reale x si y stiind ca
x²+y²-6x+4y+13=0

Answer 1

[tex]x^2+y^2-6x+4y+13=0\\ x^2-6x+9+y^2+4y+4=0\\ (x-3)^2+(y+2)^2=0\\ (x-3)^2\geq0\\ (y+2)^2\geq0\\ ......................\\ (x-3)^2+(y+2)^2\geq0\\ Dar:(x-3)^2+(y+2)^2=0=\ \textgreater \ \left \{ {{x-3=0} \atop {y+2=0}} \right.=\ \textgreater \ \left \{ {{x=3} \atop {y=-2}} \right. [/tex]