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x-x-1-x-1-x-2-find-x-




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

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