Question

Cum aflu x și y din ecuația

Answer 1

$x^2+y^2-6\sqrt 3 x+4\sqrt 5 y + 47 = 0 \\ x^2-6\sqrt 3x+(3\sqrt 3)^2 + y^2+4\sqrt 5+(2\sqrt5)^2 = 0 \\ (x-3\sqrt 3)^2+ (y+2\sqrt 5)^2 = 0 \\ \\ x = 3\sqrt 3,\quad y = -2\sqrt 5.$

Answer 2

$6 \sqrt{3} x = 2 \times 3 \sqrt{3} \: x = 2 \times \sqrt{ {3}^{2} \times 3} \: x = 2 \times \sqrt{27} \: x$

$4 \sqrt{5} \: y = 2 \times 2 \sqrt{5} \: y = 2 \times \sqrt{ {2}^{2} \times 5 } \: y = 2 \sqrt{20} \: y$

${x}^{2} + {y}^{2} - 6 \sqrt{3} \: x + 4 \sqrt{5} \: y + 47 = {x}^{2} - 2 \times \sqrt{27} \: x + {( \sqrt{27} })^{2} + {y}^{2} + 2 \times \sqrt{20} \: y + {( \sqrt{20} )}^{2} = {(x - \sqrt{27} )}^{2} + {(y + \sqrt{20} )}^{2} = 0$

${(x - \sqrt{27} )}^{2} \geqslant 0 \\ {(y + \sqrt{20} )}^{2} \geqslant 0$

=> ca egalitatea va fi numai atunci cand

$x = \sqrt{27} = 3 \sqrt{3} \: \: si \: y = - \sqrt{20} = - 2 \sqrt{5}$