Question

$x^{2}+10x+4 y^{2}-20x+50 \geq 0, pentru orice x,y apartine lui R$
Va roog!

Answer 1

$(x^2+10x+5^2)+(4y^2-20y+5^2)=(x+5)^2+(2y-5)^2 \geq 0$

Answer 2

 Fie $f(x)=x^2+10x+4y^2-20y+50$.
Avem $\Delta_f=100-4(4y^2-20y+50)=100-16y^2+80y-200=-(16y^2-80y+100)=-(4y-10)^2\le0$ .
Deci $f(x)\ge0$.