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Question Number 215564 by Wuji last updated on 10/Jan/25

$$\mathbf{given};\mathbf{x}\sqrt{\mathbf{y}}+\mathbf{y}\sqrt{\mathbf{x}}=630$$$$\mathbf{y}\sqrt{\mathbf{y}}+\mathbf{x}\sqrt{\mathbf{x}}=604$$$$\mathbf{find}\mathbf{x}\mathbf{and}\mathbf{y}$$

Answered by Frix last updated on 10/Jan/25

$$\sqrt{x}=a+b\mathrm{i}\wedge \sqrt{y}=a-b\mathrm{i};b⩾0$$$$2a^3+2{ab}^2=\alpha A$$$$2a^3-6{ab}^2=\beta B$$$$3A+B\Rightarrow a=\frac{{(3\alpha +\beta )}^{\frac{1}{3}}}{2}$$$$\mathrm{Inserting}\mathrm{in}A\mathrm{or}B\Rightarrow$$$$b=\frac{{(\alpha -\beta )}^{\frac{1}{2}}}{2{(3\alpha +\beta )}^{\frac{1}{6}}}$$$$x={(a+b\mathrm{i})}^2=\frac{\alpha +\beta }{2{(3\alpha +\beta )}^{\frac{1}{3}}}+\frac{{(\alpha -\beta )}^{\frac{1}{2}}{(3\alpha +\beta )}^{\frac{1}{6}}}{2}\mathrm{i}$$$$y={(a-b\mathrm{i})}^2=\frac{\alpha +\beta }{2{(3\alpha +\beta )}^{\frac{1}{3}}}-\frac{{(\alpha -\beta )}^{\frac{1}{2}}{(3\alpha +\beta )}^{\frac{1}{6}}}{2}\mathrm{i}$$

Commented by Frix last updated on 11/Jan/25

$$\mathrm{For}\alpha ,\beta ⩾0\mathrm{we}\mathrm{get}$$$$x=\alpha {(3\alpha +\beta )}^{-\frac{1}{3}}{\mathrm{e}}^{\mathrm{i}{\cos }^{-1}\frac{\alpha +\beta }{2\alpha }}$$$$y=\alpha {(3\alpha +\beta )}^{-\frac{1}{3}}{\mathrm{e}}^{-\mathrm{i}{\cos }^{-1}\frac{\alpha +\beta }{2\alpha }}$$$$\mathrm{which}\mathrm{look}\mathrm{a}\mathrm{bit}\mathrm{more}\mathrm{friendly}...$$

Answered by TonyCWX08 last updated on 10/Jan/25

$$u^{2}v+v^{2}u=630...E_1$$$$v^3+u^3=604...E_2$$$$$$$$3E_1+E_2$$$$u^3+3u^{2}v+3v^{2}u+v^3=2494$$$${(u+v)}^3=2494$$$$u+v=\sqrt[3]{2494}$$$$u=\sqrt[3]{2494}-v$$$$$$$$v^3+{(\sqrt[3]{2494}-v)}^3=604$$$$3\sqrt[3]{2494}{v}^2-3\sqrt[3]{{2494}^2}v+2494=604$$$$v=6.780610857\pm 0.693206253i$$$$u=6.780610858\mp 0.693206253i$$$$$$$$x={(6.780610857\pm 0.693206253i)}^2=45.49614868\pm 9.40072369i$$$$y={(6.780610858\mp 0.693206253i)}^2=45.4961487\mp 9.400723692i$$

Commented by Frix last updated on 11/Jan/25

$$\mathrm{I}\mathrm{prefer}\mathrm{exact}\mathrm{solutions}.\mathrm{If}\mathrm{I}\mathrm{want}\mathrm{approximations}$$$$\mathrm{I}\mathrm{might}\mathrm{as}\mathrm{well}\mathrm{use}\mathrm{a}\mathrm{good}\mathrm{calculator}...$$

Commented by TonyCWX08 last updated on 11/Jan/25

$$Okay$$$$I^{\prime }lleditthesolutionsothattheyareexact.$$

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