Question-219899

Question Number 219899 by MrGaster last updated on 03/May/25

Answered by Frix last updated on 03/May/25

y=((10−x^3 )/(9x^2 )) ⇒ x^9 −201x^6 +25x^3 +125=0 (x−1)(x^2 −5x−5)(x^2 +x+1)(x^4 +5x^3 +30x^2 −25x+25)=0 x=y=1 x=(5/2)±((3(√5))/2) ∧ y=(1/2)∓((√5)/2) x=y=−(1/2)±((√3)/2)i x=−((5−3(√5))/4)±(((5−3(√5))(√3))/4)i ⇒ y=−((1+(√5))/4)∓(((1+(√5))(√3))/4)i x=−((5+3(√5))/4)±(((5+3(√5))(√3))/)4i ⇒ y=−((1−(√5))/4)±(((1−(√5))(√3))/4)i

$${y}=\frac{\mathrm{10}−{x}^{\mathrm{3}} }{\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{9}} −\mathrm{201}{x}^{\mathrm{6}} +\mathrm{25}{x}^{\mathrm{3}} +\mathrm{125}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{5}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{30}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{25}\right)=\mathrm{0} \\ $$$${x}={y}=\mathrm{1} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\wedge\:{y}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}={y}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${x}=−\frac{\mathrm{5}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\pm\frac{\left(\mathrm{5}−\mathrm{3}\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{i}\:\Rightarrow\:{y}=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\mp\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{i} \\ $$$${x}=−\frac{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}}\pm\frac{\left(\mathrm{5}+\mathrm{3}\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{3}}}{}\mathrm{4i}\:\Rightarrow\:{y}=−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\pm\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{i} \\ $$

Answered by mr W last updated on 03/May/25

(((i))/((ii))) ⇒((x^3 (1+9(y/x)))/(y^3 (1+(x/y))))=((10)/2) let t=(y/x) ((1+9t)/(t^3 (1+(1/t))))=5 5t^3 +5t^2 −9t−1=0 (t−1)(5t^2 +10t+1)=0 ⇒t=1, −1±(2/( (√5))) x^3 (1+9t)=10 ⇒x=(((10)/(1+9t)))^(1/3) y^3 (1+(1/t))=2 ⇒y=(((2t)/(1+t)))^(1/3) totally 3 solutions for x, y ∈R.

$$\frac{\left({i}\right)}{\left({ii}\right)}\:\Rightarrow\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{9}\frac{{y}}{{x}}\right)}{{y}^{\mathrm{3}} \left(\mathrm{1}+\frac{{x}}{{y}}\right)}=\frac{\mathrm{10}}{\mathrm{2}} \\ $$$${let}\:{t}=\frac{{y}}{{x}} \\ $$$$\frac{\mathrm{1}+\mathrm{9}{t}}{{t}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}=\mathrm{5} \\ $$$$\mathrm{5}{t}^{\mathrm{3}} +\mathrm{5}{t}^{\mathrm{2}} −\mathrm{9}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{10}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1},\:−\mathrm{1}\pm\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${x}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{9}{t}\right)=\mathrm{10}\:\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{10}}{\mathrm{1}+\mathrm{9}{t}}} \\ $$$${y}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)=\mathrm{2}\:\Rightarrow{y}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}}} \\ $$$${totally}\:\mathrm{3}\:{solutions}\:{for}\:{x},\:{y}\:\in{R}. \\ $$

Commented by Frix last updated on 03/May/25

$$\mathrm{For}\:{x},\:{y}\:\in\mathbb{C}\:\mathrm{there}\:\mathrm{are}\:\mathrm{9}\:\mathrm{pairs}\:\left({x},\:{y}\right) \\ $$

Commented by mr W last updated on 03/May/25

$${yes} \\ $$

Answered by SdC355 last updated on 03/May/25

y=((10−x^3 )/(9x^2 )) (((10−x^3 )/(9x^2 )))^3 +x(((10−x^3 )/(9x^2 )))^2 =2 x^2 +(5/x)=6x x= 1 , ((5±3(√5))/2)

$${y}=\frac{\mathrm{10}−{x}^{\mathrm{3}} }{\mathrm{9}{x}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{10}−{x}^{\mathrm{3}} }{\mathrm{9}{x}^{\mathrm{2}} }\right)^{\mathrm{3}} +{x}\left(\frac{\mathrm{10}−{x}^{\mathrm{3}} }{\mathrm{9}{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{5}}{{x}}=\mathrm{6}{x} \\ $$$${x}=\:\mathrm{1}\:,\:\frac{\mathrm{5}\pm\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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