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a-x-a-y-2a-a-x-a-y-2a-a-R-




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

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