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Question Number 48204 by ajfour last updated on 20/Nov/18

2(x^4 −2x^2 +3)(y^4 −3y^2 +4)=7  Find (x,y) .

$$\mathrm{2}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\left({y}^{\mathrm{4}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{7} \\ $$$${Find}\:\left({x},{y}\right)\:. \\ $$

Answered by ajfour last updated on 20/Nov/18

2[(x^2 −1)^2 +2][(y^2 −(3/2))^2 +(7/4)]= 7  ⇒  x = ±1,  y = ±(√(3/2))       four points.

$$\mathrm{2}\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\right]\left[\left({y}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{4}}\right]=\:\mathrm{7} \\ $$$$\Rightarrow\:\:{x}\:=\:\pm\mathrm{1},\:\:{y}\:=\:\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:{four}\:{points}. \\ $$

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