Question

Aflati numerele a si b
a^2 +b^2+5=2a+4b

Answer 1

   
[tex]a^2 +b^2+5=2a+4b \\ a^2 +b^2+5-2a-4b=0 \\ a^2 -2a+b^2-4b+5=0 \\ a^2 -2a+b^2-4b+1+4=0 \\ a^2 -2a + 1 +b^2-4b+4=0 \\ (a^2 -2a + 1) +(b^2-4b+4)=0 \\ (a - 1)^2 + (b-2)^2 = 0 \\ \\ \Longrightarrow ~~(a - 1)^2 =0 ~~~si~~~(b-2)^2 = 0 \\ \\ (a - 1)^2 =0~~\Longrightarrow ~~ a-1 = 0~~\Longrightarrow ~~ \boxed{a = 1 } ~~~\texttt{(radacina dubla)}\\ \\ (b - 2)^2 =0~~\Longrightarrow ~~ b-2 = 0~~\Longrightarrow ~~ \boxed{b = 2 } ~~~\texttt{(radacina dubla)}[/tex]