Question

решите систему x^2+y^2=40 xy=-12

Answer 1

x=-12/y,    144/y^2+y^2=40,   144+y^4=40y^2,    y^2=t=>   t^2-40t+144=0, по правилу виета t1=36, t2=4 => y1=6, y2=-6, y3=2, y4=-2.  x1=-2, x2=2, x3=-6, x4=6

Answer 2

$displaystyleleft { {{x^2+y^2=40} atop {xy=-12}} right.~~Leftrightarrow~~left { {{(x-y)^2+2xy=40} atop {xy=-12}} right.\\\left { {{(x-y)^2-24=40} atop {xy=-12}} right.~~Leftrightarrow~~left { {{(x-y)^2=64} atop {xy=-12}} right.$

$displaystyle1)~left { {{x-y=8} atop {xy=-12}} right.~~Leftrightarrow~~left { {{x=y+8} atop {(y+8)y=-12}} right.\\\~~~~~left { {{x=y+8} atop {y^2+8y+12=0}} right.~~Leftrightarrow~~left { {{x=y+8} atop {(y+6)(y+2)=0}} right.\\\~~~~~left { {{x_1=2;~x_2=6} atop {y_1=-6;~y_2=-2}} right.$

$displaystyle2)~left { {{x-y=-8} atop {xy=-12}} right.~~Leftrightarrow~~left { {{x=y-8} atop {(y-8)y=-12}} right.\\\~~~~left { {{x=y-8} atop {y^2-8y+12=0}} right.~~Leftrightarrow~~left { {{x=y-8} atop {(y-6)(y-2)=0}} right.\\\~~~~~left { {{x_3=-2;~x_4=-6} atop {y_3=6;~y_4=2}} right.$

Ответ : (2;-6) , (6;-2) , (-2;6) , (-6;2)