Risolvi x^2+frac{sqrt{2}}{4}xy+y^2=frac{7sqrt{2}}{8}

$${ x }^{ 2 } + \frac{ \sqrt{ 2 } }{ 4 } xy+ { y }^{ 2 } = \frac{ 7 \sqrt{ 2 } }{ 8 }$$

Trova x (soluzione complessa)

$$x=\frac{\sqrt{-62y^{2}+56\sqrt{2}}-\sqrt{2}y}{8}$$

$$x=\frac{-\sqrt{-62y^{2}+56\sqrt{2}}-\sqrt{2}y}{8}$$

Trova y (soluzione complessa)

$$y=\frac{\sqrt{-62x^{2}+56\sqrt{2}}-\sqrt{2}x}{8}$$

$$y=\frac{-\sqrt{-62x^{2}+56\sqrt{2}}-\sqrt{2}x}{8}$$

Trova x

$$x=\frac{\sqrt{-62y^{2}+56\sqrt{2}}-\sqrt{2}y}{8}$$

$$x=\frac{-\sqrt{-62y^{2}+56\sqrt{2}}-\sqrt{2}y}{8}\text{, }|y|\leq \frac{2\sqrt{217}\sqrt[4]{2}}{31}$$

Trova y

$$y=\frac{\sqrt{-62x^{2}+56\sqrt{2}}-\sqrt{2}x}{8}$$

$$y=\frac{-\sqrt{-62x^{2}+56\sqrt{2}}-\sqrt{2}x}{8}\text{, }|x|\leq \frac{2\sqrt{217}\sqrt[4]{2}}{31}$$