Question

система уравнений x2+y2=20 xy=-8​

Answer 1

$\left \{ {\bigg{x^{2} + y^{2} = 20} \atop \bigg{xy = -8 \ \ \ \ \ }} \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x^{2} + y^{2} + 2xy - 2xy = 20} \atop \bigg{xy = -8 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \\\left \{ {\bigg{\ (x + y)^{2} - 2xy = 20} \atop \bigg{xy = -8 \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \ \ \ \ \ \left \{ {\bigg{\ (x+y)^{2} + 16 = 20} \atop \bigg{xy = -8 \ \ \ \ \ \ \ \ \ \ \ \ }} \right.$

$\left \{ {\bigg{\ (x+y)^{2} = 4} \atop \bigg{xy = -8 \ \ \ }} \right. \ \ \ \ \ \ \ \ \ \ \ \ \left[\begin{array}{ccc}\left \{ {\bigg{x+y = 2} \atop \bigg{xy = -8 \ }} \right. \ \ \\ \left \{ {\bigg{x+y = -2} \atop \bigg{xy = -8 \ \ \ }} \right.\\ \end{array}\right$

Решим полученные две системы уравнений, используя теорему Виета: итак, корнями первой системы будут $x_{1} = 4; \ y_{1} = -2; \ x_{2} = -2; \ y_{2} = 4$, а корнями второй системы будут $x_{3} = 2; \ y_{3} = -4; \ x_{4} = -4; \ y_{4} = 2$

Ответ: $(4; -2), \ (-2; 4), \ (2; -4), \ (-4; 2)$