(x + y)^2 = x^2 + 2xy + y^2 = 19 + 2*3 = 25
1)
{ x + y = -5
{ xy = 3
По теореме Виета x и y - корни уравнения t^2 + 5t + 3 = 0
D = 5^2 - 4*3 =25 - 12 = 13
x1 = (-5 - √13)/2; y1 = -5 - x1 = (-5 + √13)/2
x2 = (-5 + √13)/2; y2 = (-5 - √13)/2
2) x + y = 5
По теореме Виета x и y - корни уравнения t^2 - 5t + 3 = 0
x3 = (5 - √13)/2; y3 = 5 - x1 = (5 + √13)/2
x4 = (5 + √13)/2; y4 = (5 - √13)/2
Ответ: ((-5 - √13)/2; (-5 + √13)/2); ((-5 + √13)/2; (-5 - √13)/2); ((5 - √13)/2; (5 + √13)/2); ((5 + √13)/2; (5 - √13)/2)
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(x + y)^2 = x^2 + 2xy + y^2 = 19 + 2*3 = 25
1)
{ x + y = -5
{ xy = 3
По теореме Виета x и y - корни уравнения t^2 + 5t + 3 = 0
D = 5^2 - 4*3 =25 - 12 = 13
x1 = (-5 - √13)/2; y1 = -5 - x1 = (-5 + √13)/2
x2 = (-5 + √13)/2; y2 = (-5 - √13)/2
2) x + y = 5
{ x + y = -5
{ xy = 3
По теореме Виета x и y - корни уравнения t^2 - 5t + 3 = 0
D = 5^2 - 4*3 =25 - 12 = 13
x3 = (5 - √13)/2; y3 = 5 - x1 = (5 + √13)/2
x4 = (5 + √13)/2; y4 = (5 - √13)/2
Ответ: ((-5 - √13)/2; (-5 + √13)/2); ((-5 + √13)/2; (-5 - √13)/2); ((5 - √13)/2; (5 + √13)/2); ((5 + √13)/2; (5 - √13)/2)