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Question Number 152840 by liberty last updated on 02/Sep/21

Answered by MJS_new last updated on 04/Sep/21

$$\mathrm{transforming}\Rightarrow$$$$x^3-3x^2+(y^2+3)x-y(3y+1)=0$$$$x^2-\frac{1}{y}x+y^2-3=0$$$$y=px\mathrm{and}\mathrm{transforming}\Rightarrow$$$$x(x^2-3x-\frac{p-3}{p^2+1})=0[x=0\Rightarrow y=0\mathrm{not}\mathrm{possible}]$$$$x^2-\frac{3p+1}{p(p^2+1)}=0$$$$\Rightarrow$$$$x^2-3x-\frac{p-3}{p^2+1}=0$$$$x^2-\frac{3p+1}{p(p^2+1)}=0$$$$\mathrm{substracting}\mathrm{and}\mathrm{solving}\mathrm{for}x\Rightarrow$$$$x=-\frac{p^2-6p-1}{3p(p^2+1)}$$$$\mathrm{inserting}\mathrm{and}\mathrm{transforming}\Rightarrow$$$$p^6-\frac{135}{26}{p}^5-\frac{159}{26}{p}^4+\frac{9}{13}{p}^3+\frac{12}{13}{p}^2+\frac{9}{26}p+\frac{1}{26}=0$$$$(p-\frac{1}{2})(p+1)(p^2-6p-1)(p^2+\frac{4}{13}p+\frac{1}{13})=0$$$$\Rightarrow$$$$p=-1\Rightarrow x=1\wedge y=-1★$$$$p=3-\sqrt{10}\Rightarrow x=y=0\mathrm{impossible}$$$$p=\frac{1}{2}\Rightarrow x=2\wedge y=1★$$$$p=3+\sqrt{10}\Rightarrow x=y=0\mathrm{impossible}$$$$p=-\frac{2}{13}+\frac{3}{13}\mathrm{i}\Rightarrow x=\frac{3}{2}-\mathrm{i}\wedge y=\frac{1}{2}\mathrm{i}★$$$$p=-\frac{2}{13}-\frac{3}{13\mathrm{i}}\Rightarrow x=\frac{3}{2}+\mathrm{i}\wedge y=-\frac{1}{2}\mathrm{i}★$$

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