Question

Determina x, y astfel incat: x²+y²-4x+6y+10=-3.

Answer 1

[tex]x^2+y^2-4x+6y+10=-3\\ x^2+y^2-4x+6y+13=0\\ (x^2-4x+4)+(y^2+6y+9)=0\\ (x-2)^2+(y+3)^2=0\\ \text{Evident,}(x-2)^2\geq 0 \text{ si } (y+3)^2\geq 0,\text{prin urmare si suma lor e }\geq 0\\ \text{Egalitatea are loc daca si numai daca :}\\ \left \{ {{x-2=0} \atop {y+3=0}} \right.\Leftrightarrow \left \{ {{x=2} \atop {y=-3}} \right. [/tex]

Answer 2

x²+y²-4x+6y+10=-3.
x²+y²-4x+6y+13=0

x²-4x+4+y²+6y+9=0
(x-2)²+(y+3)²=0  (1)
dar

(x-2)²≥0
si
(y+3)²≥0

atunci (1)⇒
(x-2)=(y+3)²=0
deci
x=2 si y=-3