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Solve-the-system-y-x-x-y-x-y-x-y-xy-Find-all-the-real-solutions-other-than-x-0-and-y-0-

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Question Number 152178 by mathdanisur last updated on 26/Aug/21
Solve the system { ((y(√x) + x(√y) = x + y)),(((√x) + (√y) = xy)) :} Find all the real solutions other than x = 0 and y = 0
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\mathrm{y}\sqrt{\mathrm{x}}\:+\:\mathrm{x}\sqrt{\mathrm{y}}\:=\:\mathrm{x}\:+\:\mathrm{y}}\\{\sqrt{\mathrm{x}}\:+\:\sqrt{\mathrm{y}}\:=\:\mathrm{xy}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{other}\:\mathrm{than} \\ $$$$\mathrm{x}\:=\:\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\mathrm{0} \\ $$
Commented by john_santu last updated on 26/Aug/21
{ (((√(xy)) ((√y)+(√x))=x+y⇒(√y)+(√x) =((x+y)/( (√(xy)))))),(((√y)+(√x) = xy)) :}
$$\begin{cases}{\sqrt{\mathrm{xy}}\:\left(\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}\right)=\mathrm{x}+\mathrm{y}\Rightarrow\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}\:=\frac{\mathrm{x}+\mathrm{y}}{\:\sqrt{\mathrm{xy}}}}\\{\sqrt{\mathrm{y}}+\sqrt{\mathrm{x}}\:=\:\mathrm{xy}}\end{cases} \\ $$
Answered by MJS_new last updated on 26/Aug/21
x(√y)+(√x)y=x+y (√x)+(√y)=xy x=p^2 ∧p≥0 y=q^2 ∧q≥0 p^2 q+pq^2 =p^2 +q^2 p+q=p^2 q^2 p^2 q+pq^2 −p^2 −q^2 =0 p^2 q^2 −p−q=0 p=u−v∧q=u+v 2(u^3 −u^2 −uv^2 −v^2 )=0 ⇒ v^2 =((u^2 (u−1))/(u+1)) u^4 −2u^2 v^2 +v^4 −2u=0 ⇒ ⇒ u(u^3 −(1/2)u^2 −u−(1/2))=0 ⇒ u=v=0 ⇒ x=y=0 u^3 −(1/2)u^2 −u−(1/2)=0 ⇒ u=(1/6)(1+((73−6(√(87))))^(1/3) +((73+6(√(87))))^(1/3) ) ⇔ u≈1.43756489708 ⇒ v≈.609074760404 ⇒ p≈.828490136678∧q≈2.04663965749 ⇒ x≈.686395906572∧y≈4.18873388759 and obviously we can exchange x and y
$${x}\sqrt{{y}}+\sqrt{{x}}{y}={x}+{y} \\ $$$$\sqrt{{x}}+\sqrt{{y}}={xy} \\ $$$$ \\ $$$${x}={p}^{\mathrm{2}} \wedge{p}\geqslant\mathrm{0} \\ $$$${y}={q}^{\mathrm{2}} \wedge{q}\geqslant\mathrm{0} \\ $$$$ \\ $$$${p}^{\mathrm{2}} {q}+{pq}^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$${p}+{q}={p}^{\mathrm{2}} {q}^{\mathrm{2}} \\ $$$$ \\ $$$${p}^{\mathrm{2}} {q}+{pq}^{\mathrm{2}} −{p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} −{p}−{q}=\mathrm{0} \\ $$$$ \\ $$$${p}={u}−{v}\wedge{q}={u}+{v} \\ $$$$ \\ $$$$\mathrm{2}\left({u}^{\mathrm{3}} −{u}^{\mathrm{2}} −{uv}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)=\mathrm{0}\:\Rightarrow\:{v}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} \left({u}−\mathrm{1}\right)}{{u}+\mathrm{1}} \\ $$$${u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +{v}^{\mathrm{4}} −\mathrm{2}{u}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow \\ $$$${u}\left({u}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} −{u}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{u}={v}=\mathrm{0}\:\Rightarrow\:{x}={y}=\mathrm{0} \\ $$$$ \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} −{u}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:{u}=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{73}−\mathrm{6}\sqrt{\mathrm{87}}}+\sqrt[{\mathrm{3}}]{\mathrm{73}+\mathrm{6}\sqrt{\mathrm{87}}}\right) \\ $$$$\Leftrightarrow\:{u}\approx\mathrm{1}.\mathrm{43756489708} \\ $$$$\Rightarrow\:{v}\approx.\mathrm{609074760404} \\ $$$$\Rightarrow\:{p}\approx.\mathrm{828490136678}\wedge{q}\approx\mathrm{2}.\mathrm{04663965749} \\ $$$$\Rightarrow\:{x}\approx.\mathrm{686395906572}\wedge{y}\approx\mathrm{4}.\mathrm{18873388759} \\ $$$$\mathrm{and}\:\mathrm{obviously}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange}\:{x}\:\mathrm{and}\:{y} \\ $$
Answered by mr W last updated on 26/Aug/21
u=(√x)+(√y) v=(√(xy)) ⇒uv=u^2 −2v ⇒u=v^2 v^3 =v^4 −2v ⇒v=0 ⇒u=0 ⇒x=y=0 or v^3 −v^2 −2=0 (1/v^2 )+(1/(2v))−(1/2)=0 (1/v)=((((√(87))/(36))+(1/4)))^(1/3) −((((√(87))/(36))−(1/4)))^(1/3) ⇒v=(1/( ((((√(87))/(36))+(1/4)))^(1/3) −((((√(87))/(36))−(1/4)))^(1/3) )) (√x) and (√y) are roots of t^2 −ut+v=0 t^2 −v^2 t+v=0 (√x), (√y)=t=(1/2)(v^2 ±(√(v^4 −4v))) ⇒x, y=(1/4)(v^2 ±(√(v^4 −4v)))^2 ≈ { ((4.188733888)),((0.686395906)) :}
$${u}=\sqrt{{x}}+\sqrt{{y}} \\ $$$${v}=\sqrt{{xy}} \\ $$$$\Rightarrow{uv}={u}^{\mathrm{2}} −\mathrm{2}{v} \\ $$$$\Rightarrow{u}={v}^{\mathrm{2}} \\ $$$${v}^{\mathrm{3}} ={v}^{\mathrm{4}} −\mathrm{2}{v} \\ $$$$\Rightarrow{v}=\mathrm{0}\:\Rightarrow{u}=\mathrm{0}\:\Rightarrow{x}={y}=\mathrm{0} \\ $$$${or} \\ $$$${v}^{\mathrm{3}} −{v}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{v}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{v}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{v}}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{87}}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{4}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{87}}}{\mathrm{36}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\Rightarrow{v}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{87}}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{4}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{87}}}{\mathrm{36}}−\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$\sqrt{{x}}\:{and}\:\sqrt{{y}}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{2}} −{ut}+{v}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −{v}^{\mathrm{2}} {t}+{v}=\mathrm{0} \\ $$$$\sqrt{{x}},\:\sqrt{{y}}={t}=\frac{\mathrm{1}}{\mathrm{2}}\left({v}^{\mathrm{2}} \pm\sqrt{{v}^{\mathrm{4}} −\mathrm{4}{v}}\right) \\ $$$$\Rightarrow{x},\:{y}=\frac{\mathrm{1}}{\mathrm{4}}\left({v}^{\mathrm{2}} \pm\sqrt{{v}^{\mathrm{4}} −\mathrm{4}{v}}\right)^{\mathrm{2}} \approx\begin{cases}{\mathrm{4}.\mathrm{188733888}}\\{\mathrm{0}.\mathrm{686395906}}\end{cases} \\ $$
Commented by mathdanisur last updated on 26/Aug/21
Thank you Ser cool
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\ $$

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