Question Number 62228 by behi83417@gmail.com last updated on 17/Jun/19
$$\begin{cases}{\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{y}}}=\mathrm{2}\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{x}}}+\sqrt{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{y}}}=\mathrm{2}\boldsymbol{\mathrm{a}}}\end{cases}\:\:\:\:\:\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}. \\ $$
Commented by mr W last updated on 18/Jun/19
$${a}>\mathrm{0} \\ $$$$−{a}\leqslant{x},{y}\leqslant{a} \\ $$$${x}={y} \\ $$$$\sqrt{{a}+{x}}+\sqrt{{a}−{x}}=\mathrm{2}{a} \\ $$$$\sqrt{{a}+{x}}=\mathrm{2}{a}−\sqrt{{a}−{x}} \\ $$$${a}+{x}=\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}\sqrt{{a}−{x}}+{a}−{x} \\ $$$${x}=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}\sqrt{{a}−{x}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −{x}=\mathrm{2}{a}\sqrt{{a}−{x}} \\ $$$$\mathrm{4}{a}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} {x}+{x}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{3}} −\mathrm{4}{a}^{\mathrm{2}} {x} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{a}\right){a}^{\mathrm{3}} \\ $$$$\Rightarrow{x}={y}=\pm\mathrm{2}{a}\sqrt{{a}\left(\mathrm{1}−{a}\right)} \\ $$$$\Rightarrow{a}\leqslant\mathrm{1} \\ $$$$\mathrm{2}{a}\sqrt{{a}\left(\mathrm{1}−{a}\right)}\leqslant{a} \\ $$$$\Rightarrow{a}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${i}.{e}.\:{solution}\:{exists}\:{only}\:{when} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\leqslant{a}\leqslant\mathrm{1}. \\ $$
Commented by behi83417@gmail.com last updated on 18/Jun/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$