Question

решите систему уравнений х квадрат+у квадрат=26 ху=-5

Answer 1

 

$x^2 + y^2 = 26\ xy = -5\\ (x + y)^2 - 2xy = 26\ (x + y)^2 +10 = 26\\ (x+y)^2 = 16\\ x+y = 4, x+y = -4\\ 1) y(4-y) = -5\ 4y - y^2 + 5 = 0\ y^2 - 4y - 5 = 0\ y_1 + y_2 = 4\ y_1 * y_2 = -5\ y_1 = -1, y_2 = 5\ x = 5, x = -1\\ 2) y(-4-y) = -5\ 4y + y^2 = 5\ y^2 + 4y - 5 = 0\ y_1 + y_2 = -4\ y_1 * y_2 = -5\ y_1 = 1, y_2 = -5\ x = -5, x = 1\\ (5 , -1), (-1, 5), (1, -5), (-5, 1)$