Question

Решите систему уравнений х^2+у^2=34, х*у=15

Answer 1

$\tt\displaystyle \left \{ {{x^2+y^2=34} \atop {xy=15}} \right. \Longrightarrow \left \{ {{x^2+y^2=34} \atop {y=\frac{15}{x}} \right.$

Подставляем.

$\tt\displaystyle x^2+(\frac{15}{x})^2=34\\\\\\x^2+\frac{225}{x^2}=34|*x^2\\\\\\x^4+225-34x^2=0\\\\x^2=t\\\\t^2-34t+225=0\\\\D=(-34)^2-4*1*225=1156-900=256\\\\\sqrt{256}=16\\\\t_1=\frac{34-16}{2}=9\\\\\\t_2=\frac{34+16}{2}=25$

x² = t

x² = 9 или x² = 25

x = ±3        x² = ±5

y₁ = 15/3 = 5

y₂ = 15/(-3) = -5

y₃ = 15/5 = 3

y₄ = 15/(-5) = -3

ответ: (3; 5); (-3; -5); (5; 3); (-5; -3)