Question

Aflati x si y numere reale, stiind ca $x^{2} +4y^{2}-6x+4y+10 \leq 0$

Answer 1

[tex] x^{2} + 4y^{2}-6x+4y+10 \leq 0 \\ x^{2} -6x +4 y^{2}+4y+9+1 \leq 0 \\ ( x^{2} -6x+9)+(4 y^{2}-4y+1 ) \leq 0 \\ (x-3)^{2} + (2y-1)^{2} \leq 0 \\ =\ \textgreater \ (x-3)^{2} + (2y-1)^{2}=0 \\ x-3=0 =\ \textgreater \ x=3 \\ 2y-1=0 =\ \textgreater \ 2y=1 =\ \textgreater \ y=1:2 =\ \textgreater \ y= \frac{1}{2} [/tex]