Question Number 191675 by MATHEMATICSAM last updated on 28/Apr/23
$$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:{x} \\ $$
Commented by mehdee42 last updated on 29/Apr/23
$${the}\:{graph}\:{of}\:{the}\:{function}\:{is}\:{as}\:{follows}.{therefore}\:.{the}\:{given}\:\: \\ $$$${equation}\:{has}\:{only}\:{one}\:{root}\:{in}\:{the}\: \\ $$$${interval}\:\left(\mathrm{1},\mathrm{2}\right) \\ $$
Commented by mehdee42 last updated on 29/Apr/23
Answered by manxsol last updated on 30/Apr/23
$${x}>\mathrm{0} \\ $$$$\frac{{x}−\mathrm{1}}{{x}}\gg\mathrm{0}\:\:\:\:\:\:+\:\:\:\:\overset{{o}} {\mathrm{0}}\:\:\:−\:\:\:\:\overset{\bullet} {\mathrm{1}}\:+ \\ $$$${x}\geqslant\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\gg\mathrm{0}\:−\:\:-\overset{\bullet} {\mathrm{1}}\:\:\:+\:\:\:\:\overset{{o}} {\mathrm{0}}\:\:\:\:−\:\:\:\overset{\bullet} {\mathrm{1}}\:\:+ \\ $$$${x}\gg\mathrm{1} \\ $$$${sol}\:{inecuacion}\:\:{x}\geqslant\mathrm{1} \\ $$$$ \\ $$
Commented by mehdee42 last updated on 28/Apr/23
$${x}\geqslant\mathrm{0}\:,\:\frac{{x}−\mathrm{1}}{{x}}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{1} \\ $$
Answered by Frix last updated on 29/Apr/23
$${x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$