{l}{x+y=x^2}{y+x^2=y^2} का समाधान करें

$$\left. \begin{array} { l } { x + y = x ^ { 2 } } \\ { y + x ^ { 2 } = y ^ { 2 } } \end{array} \right.$$

x, y के लिए हल करें

$$x=0\text{, }y=0$$

$$x=\frac{2\left(\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+1\right)}{3}\approx 2.246979604\text{, }y=\frac{2\left(\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+14\left(\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})\right)^{2}-1\right)}{9}\approx 2.801937736$$

$$x=\frac{-\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-\sqrt{21}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+2}{3}\approx -0.801937736\text{, }y=-\frac{\left(-\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-\sqrt{21}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+2\right)\left(\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+\sqrt{21}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})+1\right)}{9}\approx 1.445041868$$

$$x=\frac{\sqrt{7}\left(\sqrt{3}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})\right)+2}{3}\approx 0.554958132\text{, }y=\frac{\left(\sqrt{7}\left(\sqrt{3}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})\right)-1\right)\left(\sqrt{7}\left(\sqrt{3}\sin(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})\right)+2\right)}{9}\approx -0.246979604$$