Question Number 56215 by necx1 last updated on 12/Mar/19
$$\sqrt{\frac{{x}}{{x}−\mathrm{1}}}+\sqrt{\frac{{x}−\mathrm{1}}{{x}}}=\mathrm{2} \\ $$$$ \\ $$$${find}\:{x} \\ $$
Commented by necx1 last updated on 12/Mar/19
$${I}\:{got}\:{confused}\:{at}\:{a}\:{point}\:{where}\:{i}\:{solved} \\ $$$${and}\:{got}\: \\ $$$$\:\:\:\:\:\:{x}={x}−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$${please}\:{help}\:{check} \\ $$
Commented by maxmathsup by imad last updated on 12/Mar/19
$${changement}\:\sqrt{\frac{{x}}{{x}−\mathrm{1}}}={t}\:{give}\:\:\frac{{x}}{{x}−\mathrm{1}}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} {x}−{t}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}\:=−{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{1}}\:\:{and}\:\left({e}\right)\:\Leftrightarrow\:{t}+\frac{\mathrm{1}}{{t}}\:=\mathrm{2}\:\Rightarrow{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{t}\:=\mathrm{0}\:\Rightarrow\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{t}=\mathrm{1}\:\Rightarrow{x}\:=\infty \\ $$$${but}\:\infty\:{is}\:{not}\:{a}\:{number}!.... \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19
$$\frac{{x}+{x}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −{x}}}=\mathrm{2} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}}=\mathrm{4} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x} \\ $$$${some}\:{thing}\:{wrong}\:{in}\:{question}... \\ $$$$ \\ $$
Commented by necx1 last updated on 12/Mar/19
$${this}\:{is}\:{exactly}\:{the}\:{same}\:{problem}\:{I} \\ $$$${encountered}.{Does}\:{it}\:{mean}\:{it}\:{cant}\:{be} \\ $$$${solve}?? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 12/Mar/19
$${t}=\frac{{x}}{{x}−\mathrm{1}} \\ $$$$\sqrt{{t}}+\frac{\mathrm{1}}{\sqrt{{t}}}=\mathrm{2} \\ $$$${z}=\sqrt{{t}} \\ $$$${z}+\frac{\mathrm{1}}{{z}}=\mathrm{2}\:\Rightarrow\:{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{z}=\mathrm{1}\:\Rightarrow\:{t}=\mathrm{1} \\ $$$$\frac{{x}}{{x}−\mathrm{1}}=\mathrm{1}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\mathrm{but}:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}}{{x}−\mathrm{1}}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}}{{x}−\mathrm{1}}\:=\mathrm{1} \\ $$
Commented by Kunal12588 last updated on 12/Mar/19
$${so}\:{what}\:{you}\:{are}\:{saying}\:{is}\:{this}\:??? \\ $$$${as}\:{x}\:{approaches}\:\infty\sqrt{\frac{{x}}{{x}−\mathrm{1}}}+\sqrt{\frac{{x}−\mathrm{1}}{{x}}} \\ $$$$=\sqrt{\mathrm{1}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}=\mathrm{1}+\mathrm{1}=\mathrm{2}\:{which}\:{satisfies}\:{the}\:{eq}^{{n}} \\ $$
Commented by MJS last updated on 12/Mar/19
$$\mathrm{yes} \\ $$
Answered by ajfour last updated on 12/Mar/19
$${squaring} \\ $$$$\frac{{x}}{{x}−\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}}\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} +\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{x}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\Rightarrow\:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}}\:=\:\mathrm{1} \\ $$$${solution} \\ $$$$\:\:\:\:\:{x}\rightarrow\pm\infty\:. \\ $$
Commented by necx1 last updated on 12/Mar/19
$${but}\:{on}\:{squaring}\:{both}\:{sides}\:{we}\:{have}\:{the} \\ $$$${RHS}=\mathrm{4} \\ $$
Commented by Kunal12588 last updated on 12/Mar/19
$${Mr}\:{ajfour}\:{is}\:{doing}\:{one}\:{step}\:{forward} \\ $$$${He}\:{is}\:{right} \\ $$
Answered by Hassen_Timol last updated on 12/Mar/19
$$ \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{when}\:{x}\:\rightarrow\:−\infty\:\:\mathrm{or}\:\:{x}\:\rightarrow\:+\infty\:\:\:{I}\:{think}... \\ $$