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Question Number 41160 by behi83417@gmail.com last updated on 02/Aug/18

Commented by behi83417@gmail.com last updated on 02/Aug/18

$$solveforx\ne y\ne z.$$

Commented by MJS last updated on 02/Aug/18

$$\mathrm{you}\mathrm{know}\mathrm{any}\mathrm{solution}?$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{there}\mathrm{is}\mathrm{one}\mathrm{but}\mathrm{I}\mathrm{could}\mathrm{be}\mathrm{wrong}$$

Commented by MJS last updated on 02/Aug/18

$$\left(1\right)z=\pm \sqrt{-x^2+xy-y^2}$$$$\left(2\right)z=\frac{y}{2}\pm \frac{\sqrt{4x^2-3y^2}}{2}$$$$\left(3\right)z=\frac{x}{2}\pm \frac{\sqrt{-3x^2+4y^2}}{2}$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{solve}$$$$\pm \sqrt{-x^2+xy-y^2}=\frac{y}{2}\pm \frac{\sqrt{4x^2-3y^2}}{2}$$$$\pm \sqrt{-x^2+xy-y^2}=\frac{x}{2}\pm \frac{\sqrt{-3x^2+4y^2}}{2}$$$$\frac{y}{2}\pm \frac{\sqrt{4x^2-3y^2}}{2}=\frac{x}{2}\pm \frac{\sqrt{-3x^2+4y^2}}{2}$$$$\mathrm{each}\mathrm{line}\mathrm{represents}4\mathrm{equations}\mathrm{but}\mathrm{all}\mathrm{of}$$$$\mathrm{them}\mathrm{have}\mathrm{got}\mathrm{the}\mathrm{only}\mathrm{solution}x=y\Rightarrow$$$$\Rightarrow x=y=z$$

Answered by MJS last updated on 02/Aug/18

$$x=y=z$$$$\mathrm{maybe}\mathrm{more}...$$$$...\mathrm{no}.$$

Commented by rahul 19 last updated on 02/Aug/18

$$\mathrm{i}\mathrm{think}\mathrm{all}\mathrm{real}\mathrm{no}.\mathrm{s}\mathrm{satisfy}\mathrm{these}\mathrm{eq}.!??$$

Commented by MJS last updated on 02/Aug/18

$$\mathrm{yes}$$

Answered by rahul 19 last updated on 02/Aug/18

$$x=y=z=k$$$$wherekϵ\mathrm{R}.$$

Answered by behi83417@gmail.com last updated on 02/Aug/18

$$2(x^2+y^2+z^2)=x^2+y^2+z^2+xy+yz+zx$$$$\Rightarrow x^2+y^2+z^2=xy+yz+zx$$$$x^2-z^2=-(x^2-z^2)+y(x-z)\Rightarrow$$$$(x-z)(2x+2z-y)=0\stackrel{x\ne z}{\Rightarrow }y=2x+2z$$$$\Rightarrow x=2z+2y,z=2x+2y\Rightarrow$$$$\Rightarrow x+y+z=0$$$${(x+y+z)}^2-2(xy+yz+zx)=xy+yz+zx$$$$\Rightarrow xy+yz+zx=0$$$$xy+z(x+y)=0\Rightarrow xy-{(x+y)}^2=0$$$$\Rightarrow x^2+xy+y^2=0\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}y$$$$z=-(x+y)=-(y+\frac{-1\pm \sqrt{5}}{2}y)=$$$$z_1=-(1-\frac{1+\sqrt{5}}{2})y=-\frac{1-\sqrt{5}}{2}y$$$$z_2=-(1-\frac{1-\sqrt{5}}{2})y=-\frac{1+\sqrt{5}}{2}y$$$$(x,y,z)=(\frac{-1\pm \sqrt{5}}{2},1,-\frac{1\mp \sqrt{5}}{2}).y$$$$x^2+y^2=y^2(1+\frac{6+2\sqrt{5}}{4})=y^2.\frac{5+\sqrt{5}}{2}$$$$z^2+xy=y^2(\frac{6-2\sqrt{5}}{4}+\frac{-2-2\sqrt{5}}{4})=y^2(1-\sqrt{5})$$$$dearMJSsir!youareright.$$$$butwhatthesolution?$$$$$$

Commented by MJS last updated on 02/Aug/18

$$\mathrm{sorry}\mathrm{but}\mathrm{this}\mathrm{is}\mathrm{wrong}$$$$(x,y,z)=(\frac{-1+\sqrt{5}}{2},1,\frac{-1-\sqrt{5}}{2})\Rightarrow$$$$x^2+y^2=\frac{5}{2}-\frac{\sqrt{5}}{2}$$$$z^2+xy=1+\sqrt{5}$$

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