Question

4x²+y²+4x+6y+10=0
x;y?

Answer 1

usor/.............
2[tex]4 x^{2} + y^{2} +4x+6y+10=0 => (2x)^{2} + y^{2} + 2*2x*1+2*3*y+9+ [/tex]+1=$(2x+1)^{2} +(Y+3)^{2} =0$
dar $(2x+1)^{2} \geq 0$, oricare ar fi x care apartine R
si $(y+3)^{2} \geq 0$, oricare ar fi y care apartine R . 
de unde rezulta ca $(2x+1)^{2} =o=> 2x+1=0=> x= \frac{-1}{2}$
$(y+3)^{2} =0=> y+3=0=> y=-3$