求证实部的绝对值相等

hbghlyj

方程组$x^2 - x y + y^2 + x - 2 y = \frac12,x^2 - y^2 = 2$的解为$(x_i,y_i)$,i=1,2,3,4,求证：$|Re(x_1)|=|Re(x_2)|=|Re(x_3)|=|Re(x_4)|$

hbghlyj

$x^2-xy+y^2+x-2y=\frac12$的式子里所有满足条件的x的绝对值小于$\sqrt2$,故方程组没有实数解,有两对共轭虚根$x_{1,2},x_{3,4}$
消y得$12 x^4 - 44 x^2 + 12 x + 57=0,\therefore x_1+x_2+x_3+x_4=0,\therefore Re(x_1)+Re(x_2)+Re(x_3)+Re(x_4)=0$
$\because Re(x_1)=Re(x_2),Re(x_3)=Re(x_4),\therefore Re(x_1)=Re(x_2)=-Re(x_3)=-Re(x_4)$
附：这个方程的根是
$x_{1,2}=-\frac{1}{2} \sqrt{\frac{7 \sqrt{13}+23}{6} }\pm \frac{1}{2} i \left(\sqrt{\frac{7 \sqrt{13}-23}{6} }-\frac{1}{\sqrt{3}}\right)$
$x_{3,4}=\frac{1}{2} \sqrt{\frac{7 \sqrt{13}+23}{6}}\pm \frac{1}{2} i \left(\sqrt{\frac{7 \sqrt{13}-23}{6}}+\frac{1}{\sqrt{3}}\right)$

kuing

回复 2# hbghlyj

`12x^4-44x^2+12x+57=12(x^2-2)^2+(2x+3)^2`

hbghlyj

好，这样也能说明方程组没有实数解咯