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Question Number 172824 by Mikenice last updated on 01/Jul/22

Commented by MJS_new last updated on 02/Jul/22

how many % can solve this? I can.

$$\mathrm{how}\:\mathrm{many}\:\%\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\mathrm{I}\:\mathrm{can}. \\ $$

Commented by cortano1 last updated on 02/Jul/22

one solution (0,0)

$${one}\:{solution}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$

Answered by mr W last updated on 02/Jul/22

let x=(√X)>0, y=(√Y)>0 2(x^2 −y^2 )y=x ...(i) (x^2 +y^2 )x=3y ...(ii) x=y=0 is solution ⇒X=Y=0 (i)×(ii): 2(x^2 −y^2 )(x^2 +y^2 )=3 ⇒x^4 −y^4 =(3/2) ...(I) (i)/(ii): ((2(x^2 −y^2 ))/(x^2 +y^2 ))=(x^2 /(3y^2 )) x^4 −5x^2 y^2 +6y^2 =0 (x^2 −2y^2 )(x^2 −3y^2 )=0 ...(II) case 1: x^2 =2y^2 ⇒4y^4 −y^4 =(3/2) ⇒y^4 =(1/2) ⇒y^2 =(1/( (√2))) ⇒Y=(1/( (√2))) ⇒x^2 =(√2) ⇒X=(√2) case 2: x^2 =3y^2 ⇒9y^4 −y^4 =(3/2) ⇒y^4 =(3/(16)) ⇒y^2 =((√3)/4) ⇒Y=((√3)/4) ⇒x^2 =((3(√3))/4) ⇒X=((3(√3))/4) summary: (X,Y)=(0,0),((√2),(1/( (√2)))),(((3(√3))/4),((√3)/4))

$${let}\:{x}=\sqrt{{X}}>\mathrm{0},\:{y}=\sqrt{{Y}}>\mathrm{0} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){y}={x}\:\:\:\:...\left({i}\right) \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){x}=\mathrm{3}{y}\:\:\:...\left({ii}\right) \\ $$$${x}={y}=\mathrm{0}\:{is}\:{solution}\:\Rightarrow{X}={Y}=\mathrm{0} \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}\:\:\:...\left({I}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\mathrm{3}{y}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \right)=\mathrm{0}\:\:\:...\left({II}\right) \\ $$$${case}\:\mathrm{1}:\:{x}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{y}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{Y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\sqrt{\mathrm{2}}\:\Rightarrow{X}=\sqrt{\mathrm{2}} \\ $$$${case}\:\mathrm{2}:\:{x}^{\mathrm{2}} =\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{y}^{\mathrm{4}} −{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{16}}\:\Rightarrow{y}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{Y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{X}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${summary}: \\ $$$$\left({X},{Y}\right)=\left(\mathrm{0},\mathrm{0}\right),\left(\sqrt{\mathrm{2}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right),\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}},\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$

Commented by Tawa11 last updated on 02/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by MJS_new last updated on 02/Jul/22

{ ((2y^(1/2) x−x^(1/2) −2y^(3/3) =0)),((x^(3/2) +yx^(1/2) −3y^(1/2) =0)) :} y=px { (((√x)(2(√p)(p−1)x+1)=0)),(((√x)((p+1)x−3(√p))=0)) :} ⇒ x_1 =0 ⇒ y_1 =0 { ((x=−(1/(2(√p)(p−1))))),((x=((3(√p))/(p+1)))) :} ⇒ p^2 −(5/6)p+(1/6)=0 ⇒ p_2 =(1/3) ⇒ x_2 =((3(√3))/4) ⇒ y_2 =((√3)/4) p_3 =(1/2) ⇒ x_3 =(√2) ⇒ y_3 =((√2)/2)

$$\begin{cases}{\mathrm{2}{y}^{\mathrm{1}/\mathrm{2}} {x}−{x}^{\mathrm{1}/\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}/\mathrm{3}} =\mathrm{0}}\\{{x}^{\mathrm{3}/\mathrm{2}} +{yx}^{\mathrm{1}/\mathrm{2}} −\mathrm{3}{y}^{\mathrm{1}/\mathrm{2}} =\mathrm{0}}\end{cases} \\ $$$${y}={px} \\ $$$$\begin{cases}{\sqrt{{x}}\left(\mathrm{2}\sqrt{{p}}\left({p}−\mathrm{1}\right){x}+\mathrm{1}\right)=\mathrm{0}}\\{\sqrt{{x}}\left(\left({p}+\mathrm{1}\right){x}−\mathrm{3}\sqrt{{p}}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow\:{y}_{\mathrm{1}} =\mathrm{0} \\ $$$$\begin{cases}{{x}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{p}}\left({p}−\mathrm{1}\right)}}\\{{x}=\frac{\mathrm{3}\sqrt{{p}}}{{p}+\mathrm{1}}}\end{cases} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{6}}{p}+\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${p}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{x}_{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow\:{y}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$${p}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}_{\mathrm{3}} =\sqrt{\mathrm{2}}\:\Rightarrow\:{y}_{\mathrm{3}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 03/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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