Question

√(x²-2x+5) + √(y²+4y+13) = 5

cum calculez?????

Answer 1

√(x²-2x+5) + √(y²+4y+13) = 5

Doar doua numere naturale adunate pot da un numar natural.

Avem doua cazuri:

(1) √(x²-2x+5) = 2             sau    (2) √(x²-2x+5) = 3
     √(y²+4y+13) = 3                         √(y²+4y+13) = 2

(1) √(x²-2x+5) = 2  => x²-2x+5  = 4 => x²-2x+1=0 => (x-1)² = 0 => x = 1
     √(y²+4y+13) = 3  => y²+4y+13 = 9 => y²+4y+4 = 0 => (y+2)² = 0 => y=-2

(2) √(x²-2x+5) = 3 => x²-2x+5 = 9 =>  x²-2x-4 = 0;
$\Delta = 4+16 = 20 \Rightarrow x_{1,2} = \frac{2\pm 2\sqrt{5} }{2} \Rightarrow x_{1,2} = 1\pm \sqrt{5$

√(y²+4y+13) = 2 =>  y²+4y+13 = 4 =>  y²+4y+9 = 0
$\Delta = 16-4\cdot 9 = 16-36 = -20 \ \textless \ 0 \Rightarrow y \in \O$

$\Rightarrow (x,y) = \Big\{(1,-2)\Big\}$