Question Number 104811 by bobhans last updated on 24/Jul/20
$${Given}\:\begin{cases}{{a}+{b}\sqrt{\mathrm{3}}−\mathrm{2}{c}\:=\:\mathrm{1}}\\{\mathrm{3}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{2}{a}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\mathrm{4}{ac}\:=\:\mathrm{5}{c}^{\mathrm{2}} }\end{cases} \\ $$$${find}\:{b} \\ $$
Answered by maths mind last updated on 24/Jul/20
$${a}^{\mathrm{2}} +\mathrm{4}{ac}=\mathrm{5}{c}^{\mathrm{2}} \Leftrightarrow\left({a}−{c}\right)\left({a}+\mathrm{5}{c}\right)=\mathrm{0} \\ $$
Answered by bemath last updated on 24/Jul/20
$${case}\left(\mathrm{1}\right)\:{a}={c} \\ $$$$\rightarrow\begin{cases}{\mathrm{3}{b}^{\mathrm{2}} =\:{a}^{\mathrm{2}} \Rightarrow{a}\:=\:\pm{b}\sqrt{\mathrm{3}}}\\{{b}\sqrt{\mathrm{3}}\:=\:\mathrm{1}+{a}\:=\:\mathrm{1}−{b}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{2}{b}\sqrt{\mathrm{3}}\:=\:\mathrm{1}\:\Rightarrow\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\:\bigstar \\ $$$${case}\left(\mathrm{2}\right)\:{a}\:=\:−\mathrm{5}{c}\: \\ $$$$\rightarrow\begin{cases}{\mathrm{3}{b}^{\mathrm{2}} \:=\:\mathrm{49}{c}^{\mathrm{2}} \Rightarrow{c}=\:\pm\frac{{b}\sqrt{\mathrm{3}}}{\mathrm{7}}}\\{{b}\sqrt{\mathrm{3}}\:=\:\mathrm{1}+\mathrm{7}{c}\:=\:\mathrm{1}\pm{b}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{2}{b}\sqrt{\mathrm{3}}\:=\:\mathrm{1}\:\Rightarrow{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\:\bigstar \\ $$$$ \\ $$