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Question Number 65457 by behi83417@gmail.com last updated on 30/Jul/19
{ (((a/b)+((b−a)/(b+a))=2(√3))),(((b/a)+((b+a)/(b−a))=3(√2))) :} [a,b∈R,a≠b]
$$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}+\frac{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}=\mathrm{2}\sqrt{\mathrm{3}}}\\{\frac{\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}}+\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}}=\mathrm{3}\sqrt{\mathrm{2}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}},\boldsymbol{\mathrm{a}}\neq\boldsymbol{\mathrm{b}}\right] \\ $$
Answered by mr W last updated on 30/Jul/19
let t=(a/b) eqn. 1 ⇒t+((1−t)/(1+t))=2(√3) ...(i) eqn. 2 ⇒(1/t)+((1+t)/(1−t))=3(√2) ...(ii) (i): t^2 −2(√3)t−(2(√3)−1)=0 t=(√3)±(√(2((√3)+1))) ←real (ii): 1−t+t^2 +t=3(√2)(t−t^2 ) (1+3(√2))t^2 −3(√2)t+1=0 t=((3(√2)±(√(2(7−6(√2)))))/(2(1+3(√2)))) ←not real ⇒no solution!
$${let}\:{t}=\frac{{a}}{{b}} \\ $$$${eqn}.\:\mathrm{1}\:\Rightarrow{t}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}=\mathrm{2}\sqrt{\mathrm{3}}\:\:\:\:\:\:…\left({i}\right) \\ $$$${eqn}.\:\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}=\mathrm{3}\sqrt{\mathrm{2}}\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right): \\ $$$${t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{t}−\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}\:\leftarrow{real} \\ $$$$ \\ $$$$\left({ii}\right): \\ $$$$\mathrm{1}−{t}+{t}^{\mathrm{2}} +{t}=\mathrm{3}\sqrt{\mathrm{2}}\left({t}−{t}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\pm\sqrt{\mathrm{2}\left(\mathrm{7}−\mathrm{6}\sqrt{\mathrm{2}}\right)}}{\mathrm{2}\left(\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}\right)}\:\leftarrow{not}\:{real} \\ $$$$ \\ $$$$\Rightarrow{no}\:{solution}! \\ $$
Commented by behi83417@gmail.com last updated on 30/Jul/19
thank you very much dear master.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$

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