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Question Number 27507 by NECx last updated on 07/Jan/18

$$2\sqrt{x}+y=9....\left(1\right)$$$$x+2\sqrt{y}=3....\left(2\right)$$$$$$$$solvethesimultaneousequation$$

Commented by Rasheed.Sindhi last updated on 08/Jan/18

$$\left(1\right)+\left(2\right):$$$$\mathrm{x}+\mathrm{y}+2(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})=12.....\left(3\right)$$$$\left(1\right)-\left(2\right):$$$$-(\mathrm{x}-\mathrm{y})+2(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})=6...\left(4\right)$$$$\mathrm{Let}\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{u}\mathrm{\&}\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}}=\mathrm{v}$$$${(\mathrm{a}+\mathrm{b})}^2-{(\mathrm{a}-\mathrm{b})}^2=4\mathrm{ab}$$$${\mathrm{u}}^2-{\mathrm{v}}^2=4\sqrt{\mathrm{xy}}$$$${}^{\bullet }\sqrt{\mathrm{xy}}=\frac{{\mathrm{u}}^2-{\mathrm{v}}^2}{4}$$$${(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})}^2={\mathrm{u}}^2$$$$\mathrm{x}+\mathrm{y}+2\sqrt{\mathrm{xy}}={\mathrm{u}}^2$$$${}^{\bullet }\mathrm{x}+\mathrm{y}={\mathrm{u}}^2-2\left(\frac{{\mathrm{u}}^2-{\mathrm{v}}^2}{4}\right)$$$$=\frac{{\mathrm{u}}^2+{\mathrm{v}}^2}{2}$$$${(\mathrm{a}-\mathrm{b})}^2={(\mathrm{a}+\mathrm{b})}^2-4\mathrm{ab}$$$${}^{\bullet }{(\mathrm{x}-\mathrm{y})}^2={\left(\frac{{\mathrm{u}}^2+{\mathrm{v}}^2}{2}\right)}^2-4{\left(\frac{{\mathrm{u}}^2-{\mathrm{v}}^2}{4}\right)}^2$$$$=\frac{{({\mathrm{u}}^2+{\mathrm{v}}^2)}^2}{4}-\frac{{({\mathrm{u}}^2-{\mathrm{v}}^2)}^2}{4}$$$$=\frac{{\mathrm{u}}^4+{2\mathrm{u}}^2{\mathrm{v}}^2+{\mathrm{v}}^4-{\mathrm{u}}^4+{2\mathrm{u}}^2{\mathrm{v}}^2-{\mathrm{v}}^4}{4}$$$${(\mathrm{x}-\mathrm{y})}^2={\mathrm{u}}^2{\mathrm{v}}^2$$$$\mathrm{x}-\mathrm{y}=\pm \mathrm{uv}$$$$\left(3\right):\mathrm{x}+\mathrm{y}+2(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}})=12$$$$\frac{{\mathrm{u}}^2+{\mathrm{v}}^2}{2}+2\mathrm{u}=12$$$${\mathrm{u}}^2+4\mathrm{u}+{\mathrm{v}}^2=24.......\left(5\right)$$$$\left(4\right):-(\mathrm{x}-\mathrm{y})+2(\sqrt{\mathrm{x}}-\sqrt{\mathrm{y}})=6$$$$-(\pm \mathrm{uv})+2\mathrm{v}=6$$$$\mp \mathrm{uv}+2\mathrm{v}=6$$$$\mathrm{v}=\frac{6}{\mp \mathrm{u}+2}$$$$\left(5\right):{\mathrm{u}}^2+4\mathrm{u}+{\left(\frac{6}{\mp \mathrm{u}+2}\right)}^2=24$$$$\left(5\right):\mathrm{u}(\mathrm{u}+4)+\frac{36}{{(\mp \mathrm{u}+2)}^2}=24$$$$\mathrm{u}(\mathrm{u}+4){(\mp \mathrm{u}+2)}^2+36=24{(\mp \mathrm{u}+2)}^2$$$$\mathrm{Continue}$$

Commented by prakash jain last updated on 08/Jan/18

$$2\sqrt{x}+y=9$$$$y_1=9-2\sqrt{x}$$$$x+2\sqrt{y}=3$$$$2\sqrt{y}=3-x$$$$\mathrm{for}\mathrm{real}y,x⩽3$$$$y_2=\frac{1}{4}{(3-x)}^2$$$$u=y_1-y_2=9-2\sqrt{x}-\frac{1}{4}{(3-x)}^2$$$$forrealsolutionx⩽3$$$$x⩽3\Rightarrow 9-2\sqrt{x}⩾9-2\sqrt{3}>5$$$$\frac{1}{4}{(3-x)}^2⩽\frac{9}{4}$$$$u⩾5-\frac{9}{4}>0\Rightarrow norealsolution$$

Commented by Rasheed.Sindhi last updated on 08/Jan/18

$$\mathrm{New}\mathrm{way}\mathrm{for}\mathrm{me}\mathrm{to}\mathrm{attack}\mathrm{such}$$$$\mathrm{problem}!$$

Commented by prakash jain last updated on 08/Jan/18

Actually finding the complex solutions is tough.

Answered by jota@ last updated on 09/Jan/18

$$thereisnotarealsolution.$$

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