Question

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

Answer 1

$x^{2} +4 y^{2} -6x+4y+10 \leq 0 <=> \\ <=>( x^{2} -6x+9)+( 4y^{2}+4y+1) \leq 0<=> \\ <=>(x-3)^{2}+(2y+1)^{2} \leq 0$.........(1)

Cum $(x-3)^{2} \geq 0$ si $(2y+1)^{2} \geq 0$, rezulta ca $(x-3)^{2} +(2y+1)^{2} \geq 0 ........(2)$.

Din relatiile (1) si (2) => $(x-3)^{2} +(2y+1)^{2} =0$,dar $(x-3)^{2} \geq 0$ si $(2y+1)^{2} \geq 0$ => $(x-3)^{2}=0 => x-3=0 =>x=3$ si $(2y+1)^{2} =0 =>2y+1=0 =>2y=-1=>y=- \frac{1}{2}$.

Solutie: $(x;y)=(3; - \frac{1}{2})$.