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Question Number 215564 by Wuji last updated on 10/Jan/25
given; x(√y)+y(√x)=630 y(√y)+x(√x)=604 find x and y
$$\boldsymbol{\mathrm{given}};\:\:\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{630} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{604} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$
Answered by Frix last updated on 10/Jan/25
(√x)=a+bi∧(√y)=a−bi; b≥0 2a^3 +2ab^2 =α A 2a^3 −6ab^2 =β B 3A+B ⇒ a=(((3α+β)^(1/3) )/2) Inserting in A or B ⇒ b=(((α−β)^(1/2) )/(2(3α+β)^(1/6) )) x=(a+bi)^2 =((α+β)/(2(3α+β)^(1/3) ))+(((α−β)^(1/2) (3α+β)^(1/6) )/2)i y=(a−bi)^2 =((α+β)/(2(3α+β)^(1/3) ))−(((α−β)^(1/2) (3α+β)^(1/6) )/2)i
$$\sqrt{{x}}={a}+{b}\mathrm{i}\wedge\sqrt{{y}}={a}−{b}\mathrm{i};\:{b}\geqslant\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{3}} +\mathrm{2}{ab}^{\mathrm{2}} =\alpha\:\:\:{A} \\ $$$$\mathrm{2}{a}^{\mathrm{3}} −\mathrm{6}{ab}^{\mathrm{2}} =\beta\:\:\:{B} \\ $$$$\mathrm{3}{A}+{B}\:\Rightarrow\:{a}=\frac{\left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:{A}\:\mathrm{or}\:{B}\:\Rightarrow \\ $$$${b}=\frac{\left(\alpha−\beta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{6}}} } \\ $$$${x}=\left({a}+{b}\mathrm{i}\right)^{\mathrm{2}} =\frac{\alpha+\beta}{\mathrm{2}\left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }+\frac{\left(\alpha−\beta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{2}}\mathrm{i} \\ $$$${y}=\left({a}−{b}\mathrm{i}\right)^{\mathrm{2}} =\frac{\alpha+\beta}{\mathrm{2}\left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }−\frac{\left(\alpha−\beta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}\alpha+\beta\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{2}}\mathrm{i} \\ $$
Commented by Frix last updated on 11/Jan/25
For α, β ≥0 we get x=α(3α+β)^(−(1/3)) e^(i cos^(−1) ((α+β)/(2α))) y=α(3α+β)^(−(1/3)) e^(−i cos^(−1) ((α+β)/(2α))) which look a bit more friendly...
$$\mathrm{For}\:\alpha,\:\beta\:\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\alpha\left(\mathrm{3}\alpha+\beta\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{e}^{\mathrm{i}\:\mathrm{cos}^{−\mathrm{1}} \:\frac{\alpha+\beta}{\mathrm{2}\alpha}} \\ $$$${y}=\alpha\left(\mathrm{3}\alpha+\beta\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{e}^{−\mathrm{i}\:\mathrm{cos}^{−\mathrm{1}} \:\frac{\alpha+\beta}{\mathrm{2}\alpha}} \\ $$$$\mathrm{which}\:\mathrm{look}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{more}\:\mathrm{friendly}… \\ $$
Answered by TonyCWX08 last updated on 10/Jan/25
u^2 v+v^2 u=630...E_1 v^3 +u^3 =604...E_2 3E_1 +E_2 u^3 +3u^2 v+3v^2 u+v^3 =2494 (u+v)^3 =2494 u+v=((2494))^(1/3) u=((2494))^(1/3) −v v^3 +(((2494))^(1/3) −v)^3 =604 3((2494))^(1/3) v^2 −3((2494^2 ))^(1/3) v+2494=604 v=6.780610857±0.693206253i u=6.780610858∓0.693206253i x=(6.780610857±0.693206253i)^2 =45.49614868±9.40072369i y=(6.780610858∓0.693206253i)^2 =45.4961487∓9.400723692i
$${u}^{\mathrm{2}} {v}+{v}^{\mathrm{2}} {u}=\mathrm{630}…{E}_{\mathrm{1}} \\ $$$${v}^{\mathrm{3}} +{u}^{\mathrm{3}} =\mathrm{604}…{E}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{3}{E}_{\mathrm{1}} +{E}_{\mathrm{2}} \\ $$$${u}^{\mathrm{3}} +\mathrm{3}{u}^{\mathrm{2}} {v}+\mathrm{3}{v}^{\mathrm{2}} {u}+{v}^{\mathrm{3}} =\mathrm{2494} \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} =\mathrm{2494} \\ $$$${u}+{v}=\sqrt[{\mathrm{3}}]{\mathrm{2494}} \\ $$$${u}=\sqrt[{\mathrm{3}}]{\mathrm{2494}}−{v} \\ $$$$ \\ $$$${v}^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{2494}}−{v}\right)^{\mathrm{3}} =\mathrm{604} \\ $$$$\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2494}}{v}^{\mathrm{2}} −\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2494}^{\mathrm{2}} }{v}+\mathrm{2494}=\mathrm{604} \\ $$$${v}=\mathrm{6}.\mathrm{780610857}\pm\mathrm{0}.\mathrm{693206253}{i} \\ $$$${u}=\mathrm{6}.\mathrm{780610858}\mp\mathrm{0}.\mathrm{693206253}{i} \\ $$$$ \\ $$$${x}=\left(\mathrm{6}.\mathrm{780610857}\pm\mathrm{0}.\mathrm{693206253}{i}\right)^{\mathrm{2}} =\mathrm{45}.\mathrm{49614868}\pm\mathrm{9}.\mathrm{40072369}{i} \\ $$$${y}=\left(\mathrm{6}.\mathrm{780610858}\mp\mathrm{0}.\mathrm{693206253}{i}\right)^{\mathrm{2}} =\mathrm{45}.\mathrm{4961487}\mp\mathrm{9}.\mathrm{400723692}{i} \\ $$
Commented by Frix last updated on 11/Jan/25
I prefer exact solutions. If I want approximations I might as well use a good calculator...
$$\mathrm{I}\:\mathrm{prefer}\:\mathrm{exact}\:\mathrm{solutions}.\:\mathrm{If}\:\mathrm{I}\:\mathrm{want}\:\mathrm{approximations} \\ $$$$\mathrm{I}\:\mathrm{might}\:\mathrm{as}\:\mathrm{well}\:\mathrm{use}\:\mathrm{a}\:\mathrm{good}\:\mathrm{calculator}… \\ $$
Commented by TonyCWX08 last updated on 11/Jan/25
Okay I′ll edit the solution so that they are exact.
$${Okay} \\ $$$${I}'{ll}\:{edit}\:{the}\:{solution}\:{so}\:{that}\:{they}\:{are}\:{exact}. \\ $$

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