Question

va rog ajutor!Determinati nr reale x si y stiin ca : x'2+4y'2-6x+4y+10=0 .. am pus '2 ceea ce inseamna la puterea a doua.. altfel nu puteam scrie..

Answer 1

[tex]x^{2}+4y^{2}-6x+4y+10=0\\10=1+9\\9\ la\ x\ si\ 1\ la\ y\\ x^{2}-6x+9=x^{2}-3x-3x+9=\\x(x-3)-3(x-3)=(x-3)(x-3)=(x-3)^{2}\\ 4y^{2}+4y+1=4y^{2}+2y+2y+1=2y(2y+1)+1(2y+1)=\\(2y+1)(2y+1)=(2y+1)^{2}\\ (x-3)^{2}+(2y+1)^{2}=0\\ Fiecare\ patrat\ perfect\ e\ pozitiv\ si\ trebuie\ sa\ dea\ 0\\ (x-3)^{2}=0\\\sqrt{(x-3)^{2}}=\sqrt{0}\\|x-3|=0\\x=3\\ (2y+1)^{2}=0\\\sqrt{(2y+1)^{2}}=\sqrt{0}\\|2y+1|=0\\2y=-1\\y=-\frac{1}{2}[/tex]