Question

СРОЧНО Решите систему уравнений x+y^2=4, xy^2=3.

Answer 1

$\left \{ {{x+y^2=4} \atop {xy^2=3}} \right. \\ \\ x+y^2=4 >> y^2=4-x \\ \\ x(4-x)=3 \\ 4x-x^2-3=0 | * (-1) \\ x^2-4x+3=0 \\ d=4 \\ \\ x_{1} = 3 \\ x_{2} = 1 \\ \\ \\ y^2 = 4-x \\ y= \sqrt{4-x} \\ \\ y_{1} = \sqrt{4-3} = \sqrt{1} = +-1 \\ y_{1} = \sqrt{4-1} = +-\sqrt{3}$

Ответ: $(3; 1), (3; -1), (1; \sqrt{3}), (1; -\sqrt{3}).$