Question

Aflati x  si y numere reale, stiind ca x²+6x+4y²-4y+10=0

Answer 1

( x  + 3 )² + ( 2y  - 1) = 0 
daca  x +3 = 0       ; x = - 3 
          2y - 1 = 0    ; y = 1 /2 

Answer 2

  
[tex]\displaystyle x^2+6x+4y^2-4y+10=0 \\ x^2+6x+4y^2-4y+9+1=0 \\ (x^2+6x + 9) + (4y^2-4y+1)=0 \\ (x^2+2 \cdot 3\cdot x + 3^2) + ((2y)^2+2 \cdot 2y \cdot (-1) +(-1)^2)=0 \\ (x+3)^2 + (2y-1)^2=0 \\ x+3 = 0 ~~\Longrightarrow~~x = \boxed{-3} ~~~(radacina ~~dubla) \\ 2y-1=0 \Longrightarrow~~ y = \boxed{\frac{1}{2}} ~~~(radacina ~~dubla) [/tex]