Question

Dacă a^2+b^2-4a+6b+13=0, aflați a+b.
*eu am descompus aşa: a^2-4a+4+b^2+6b+9=0=>(a-2)^2 + (b+3)^2=0 , dar de aici nu mai știu*

Answer 1

$a^2+b^2-4a+6b+13=0 \\ (a^2-4a+4)+(b^2+6b+9)=0 \\ (a-2)^2+(b+3)^2=0 \\ \\ (a-2)^2 \geq 0~si~(b+3)^2 \geq 0~\Rightarrow (a-2)^2+(b+3)^2 \geq 0,~dar~ \\ (a-2)^2+(b+3)^2=0 \Rightarrow~(a-2)^2=0~si~(b+3)^2=0. \\ \\ (a-2)^2=0 \Rightarrow a=2. \\ (b+3)^2 =0 \Rightarrow b=-3. \\ \\ a+b=2+(-3)=-1.$