Question Number 89461 by student work last updated on 17/Apr/20
Answered by mahdi last updated on 17/Apr/20
$$\mathrm{A}=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}+\frac{\mathrm{y}}{…}}}\Rightarrow\mathrm{A}=\mathrm{x}+\frac{\mathrm{y}}{\mathrm{A}}\Rightarrow \\ $$$$\mathrm{A}^{\mathrm{2}} =\mathrm{Ax}+\mathrm{y}\Rightarrow\mathrm{A}^{\mathrm{2}} −\mathrm{xA}−\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{A}=\frac{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4y}}}{\mathrm{2}}\:\mathrm{or}\:\frac{\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4y}}}{\mathrm{2}} \\ $$
Commented by mahdi last updated on 17/Apr/20
$$\left.\mathrm{a}\left.\right)\left.\mathrm{6}\left.\:\:\:\:\:\:\:\mathrm{b}\right)\mathrm{2}−\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{c}\right)\mathrm{4}\:\:\:\:\:\:\:\mathrm{d}\right)\mathrm{4} \\ $$
Answered by $@ty@m123 last updated on 17/Apr/20
$$\left({a}\right)\:{x}=\mathrm{5}+\frac{\mathrm{6}}{{x}}\:..\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} =\mathrm{5}{x}+\mathrm{6} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{6}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{6},\:−\mathrm{1} \\ $$$${but}\:{x}=−\mathrm{1}\:{is}\:{inadmissible}. \\ $$$$\therefore\:{x}=\mathrm{6}. \\ $$$$\left({b}\right)\:{x}=\mathrm{3}−\frac{\mathrm{2}}{{x}} \\ $$$$…. \\ $$$$…. \\ $$$$\left({c}\right)\:{x}=\mathrm{3}+\frac{\mathrm{4}}{{x}} \\ $$$$….. \\ $$$$….. \\ $$$$\left({d}\right)\:{x}=\mathrm{2}−\frac{\mathrm{8}}{{x}} \\ $$$$….. \\ $$$$….. \\ $$