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Question Number 65495 by behi83417@gmail.com last updated on 30/Jul/19
{ (((a/b)+(b/(a+b))=(√3))),(((b/a)+(a/(a+b))=(√2))) :} [a,b∈R]
$$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{3}}}\\{\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}=\sqrt{\mathrm{2}}}\end{cases}\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}\right] \\ $$
Answered by MJS last updated on 31/Jul/19
b=at { ((((t^2 +t+1)/(t(t+1)))=(√3))),((((t^2 +t+1)/(t+1))=(√2))) :} { ((t^2 +t−((1+(√3))/2)=0)),((t^2 +(1−(√2))t+1−(√2)=0)) :} { ((t=−(1/2)±((√(3+2(√3)))/2))),((t=−((1−(√2))/2)±((√(−1+2(√2)))/2))) :} ⇒ no solution
$${b}={at} \\ $$$$\begin{cases}{\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)}=\sqrt{\mathrm{3}}}\\{\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}+\mathrm{1}}=\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\begin{cases}{{t}^{\mathrm{2}} +{t}−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{0}}\\{{t}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){t}+\mathrm{1}−\sqrt{\mathrm{2}}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{t}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{2}}}\\{{t}=−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$

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