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Question Number 62275 by behi83417@gmail.com last updated on 18/Jun/19

{ ((a(√x)+b(√y)=2(√(ab)))),((x(√a)+y(√b)=2(√(ab)))) :} a,b∈R^+

$$\begin{cases}{\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{y}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\\{\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{a}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{b}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\end{cases}\:\:\:\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$

Answered by mr W last updated on 19/Jun/19

let X=(√x),Y=(√y),A=(√a),B=(√b) A^2 X+B^2 Y=2AB ⇒Y=((2AB−A^2 X)/B^2 ) AX^2 +BY^2 =2AB AX^2 +B(((2AB−A^2 X)^2 )/B^4 )=2AB (A^3 +B^3 )X^2 −4A^2 BX+2B^2 (2A−B^2 )=0 ⇒X=((2A^2 B±B^2 (√(2(A^3 +B^3 )−4AB)))/(A^3 +B^3 )) ⇒x={((2a(√b)±b(√(2(a(√a)+b(√b))−4(√(ab)))))/(a(√a)+b(√b)))}^2 similarly (A^3 +B^3 )Y^2 −4AB^2 Y+2A^2 (2B−A^2 )=0 ⇒Y=((2AB^2 ±A^2 (√(2(A^3 +B^2 )−4AB)))/(A^3 +B^3 )) ⇒y={((2(√a)b±a(√(2(a(√a)+b(√b))−4(√(ab)))))/(a(√a)+b(√b)))}^2

$${let}\:{X}=\sqrt{{x}},{Y}=\sqrt{{y}},{A}=\sqrt{{a}},{B}=\sqrt{{b}} \\ $$$${A}^{\mathrm{2}} {X}+{B}^{\mathrm{2}} {Y}=\mathrm{2}{AB} \\ $$$$\Rightarrow{Y}=\frac{\mathrm{2}{AB}−{A}^{\mathrm{2}} {X}}{{B}^{\mathrm{2}} } \\ $$$${AX}^{\mathrm{2}} +{BY}^{\mathrm{2}} =\mathrm{2}{AB} \\ $$$${AX}^{\mathrm{2}} +{B}\frac{\left(\mathrm{2}{AB}−{A}^{\mathrm{2}} {X}\right)^{\mathrm{2}} }{{B}^{\mathrm{4}} }=\mathrm{2}{AB} \\ $$$$\left({A}^{\mathrm{3}} +{B}^{\mathrm{3}} \right){X}^{\mathrm{2}} −\mathrm{4}{A}^{\mathrm{2}} {BX}+\mathrm{2}{B}^{\mathrm{2}} \left(\mathrm{2}{A}−{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{X}=\frac{\mathrm{2}{A}^{\mathrm{2}} {B}\pm{B}^{\mathrm{2}} \sqrt{\mathrm{2}\left({A}^{\mathrm{3}} +{B}^{\mathrm{3}} \right)−\mathrm{4}{AB}}}{{A}^{\mathrm{3}} +{B}^{\mathrm{3}} } \\ $$$$\Rightarrow{x}=\left\{\frac{\mathrm{2}{a}\sqrt{{b}}\pm{b}\sqrt{\mathrm{2}\left({a}\sqrt{{a}}+{b}\sqrt{{b}}\right)−\mathrm{4}\sqrt{{ab}}}}{{a}\sqrt{{a}}+{b}\sqrt{{b}}}\right\}^{\mathrm{2}} \\ $$$$ \\ $$$${similarly} \\ $$$$\left({A}^{\mathrm{3}} +{B}^{\mathrm{3}} \right){Y}^{\mathrm{2}} −\mathrm{4}{AB}^{\mathrm{2}} {Y}+\mathrm{2}{A}^{\mathrm{2}} \left(\mathrm{2}{B}−{A}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{Y}=\frac{\mathrm{2}{AB}^{\mathrm{2}} \pm{A}^{\mathrm{2}} \sqrt{\mathrm{2}\left({A}^{\mathrm{3}} +{B}^{\mathrm{2}} \right)−\mathrm{4}{AB}}}{{A}^{\mathrm{3}} +{B}^{\mathrm{3}} } \\ $$$$\Rightarrow{y}=\left\{\frac{\mathrm{2}\sqrt{{a}}{b}\pm{a}\sqrt{\mathrm{2}\left({a}\sqrt{{a}}+{b}\sqrt{{b}}\right)−\mathrm{4}\sqrt{{ab}}}}{{a}\sqrt{{a}}+{b}\sqrt{{b}}}\right\}^{\mathrm{2}} \\ $$

Commented by behi83417@gmail.com last updated on 20/Jun/19

perfect sir.thank you so much dear master.

$$\mathrm{perfect}\:\mathrm{sir}.\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$

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