Question

Determinaţi numerele reale x
şi y astfel încât x²+4y²+10x+4y+26=0 .

Answer 1

x²+(2y)²+2·5·x+2·1·2y+1+25=0
x²+2·5·x+25+(2y)²+2·1·2y+1=0
x²+2·5·x+5²+(2y)²+2·1·2y+1²=0
(x+5)²+(2y+1)²=0
(x+5)²=0 si (2y+1)²=0
√(x+5)²=√0 si √(2y+1)²=√0
|x+5|=0 si |2y+1|=0
x+5=0 si 2y+1=0
x=-5 si y=-½

Answer 2

$\it{x^2+4y^2+10x+4y+26=0\Leftrightarrow (x^2+10x+25)+(4y^2+4y+1)=0}$

[tex]\it\Leftrightarrow (x+5)^2+(2y+1)^2=0 \Leftrightarrow \begin{cases}\it x+5=0 \Leftrightarrow x=-5 \\ \\ \it2y+1=0 \Leftrightarrow y =-\dfrac{1}{2}\end{cases}[/tex]