Question

Aflati numerele reale a si b astfel incat sa fie indeplinita conditia : a²+b²-4a-10b+29≤0

Answer 1

$a^2+b^2-4a-10b+29 \leq 0 \Leftrightarrow \\ \\ \Leftrightarrow (a^2-4a+4)+(b^2-10b+25) \leq 0 \Leftrightarrow \\ \\ \Leftrightarrow (a-2)^2+(b-5)^2 \leq 0.~~~~(1) \\ \\ Dar~(a-2)^2 \geq 0~si~(b-5)^2 \geq 0 \Rightarrow~(a-2)^2+(b-5)^2 \geq 0.~~~~(2) \\ \\ Din~(1)~si~(2)~rezulta~(a-2)^2+(b-5)^2=0 \Rightarrow a-2=0~si~b-5=0. \\ \\ Asadar~a=2~si~b=5.$