(√(x/(x−1)))+(√((x−1)/x))=2 find x


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Question Number 56215 by necx1 last updated on 12/Mar/19

$\sqrt{\frac{x}{x - 1}} + \sqrt{\frac{x - 1}{x}} = 2$ $f i n d x$
Commented by necx1 last updated on 12/Mar/19

$I g o t c o n f u s e d a t a p o i n t w h e r e i s o l v e d$ $a n d g o t$ $x = x - 1$ $p l e a s e h e l p c h e c k$
Commented by maxmathsup by imad last updated on 12/Mar/19

$c h a n g e m e n t \sqrt{\frac{x}{x - 1}} = t g i v e \frac{x}{x - 1} = t^{2} \Rightarrow x = t^{2} x - t^{2} \Rightarrow (1 - t^{2}) x = - t^{2} \Rightarrow$ $x = \frac{t^{2}}{t^{2} - 1} a n d (e) \Leftrightarrow t + \frac{1}{t} = 2 \Rightarrow t^{2} + 1 - 2 t = 0 \Rightarrow {(t - 1)}^{2} = 0 \Rightarrow t = 1 \Rightarrow x = \infty$ $b u t \infty i s n o t a n u m b e r! . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19

	$\frac{x + x - 1}{\sqrt{x^{2} - x}} = 2$ $\frac{4 x^{2} - 4 x + 1}{x^{2} - x} = 4$ $4 x^{2} - 4 x + 1 = 4 x^{2} - 4 x$ $s o m e t h i n g w r o n g i n q u e s t i o n . . .$
	Commented by necx1 last updated on 12/Mar/19

	$t h i s i s e x a c t l y t h e s a m e p r o b l e m I$ $e n c o u n t e r e d . D o e s i t m e a n i t c a n t b e$ $s o l v e ? ?$
	Answered by MJS last updated on 12/Mar/19

	$t = \frac{x}{x - 1}$ $\sqrt{t} + \frac{1}{\sqrt{t}} = 2$ $z = \sqrt{t}$ $z + \frac{1}{z} = 2 \Rightarrow z^{2} - 2 z + 1 = 0 \Rightarrow z = 1 \Rightarrow t = 1$ $\frac{x}{x - 1} = 1 \Rightarrow no real solution$ $but : \lim_{x \to \infty} \frac{x}{x - 1} = \lim_{x \to - \infty} \frac{x}{x - 1} = 1$
	Commented by Kunal12588 last updated on 12/Mar/19

	$s o w h a t y o u a r e s a y i n g i s t h i s ? ? ?$ $a s x a p p r o a c h e s \infty \sqrt{\frac{x}{x - 1}} + \sqrt{\frac{x - 1}{x}}$ $= \sqrt{1} + \sqrt{\frac{1}{1}} = 1 + 1 = 2 w h i c h s a t i s f i e s t h e {e q}^{n}$
	Commented by MJS last updated on 12/Mar/19

	$yes$
	Answered by ajfour last updated on 12/Mar/19

	$s q u a r i n g$ $\frac{x}{x - 1} + \frac{x - 1}{x} = 2$ $\Rightarrow x^{2} + {(x - 1)}^{2} = 2 x (x - 1)$ $\Rightarrow \Rightarrow \frac{2 x^{2} - 2 x + 1}{2 x^{2} - 2 x} = 1$ $s o l u t i o n$ $x \to \pm \infty .$
	Commented by necx1 last updated on 12/Mar/19

	$b u t o n s q u a r i n g b o t h s i d e s w e h a v e t h e$ $R H S = 4$
	Commented by Kunal12588 last updated on 12/Mar/19

	$M r a j f o u r i s d o i n g o n e s t e p f o r w a r d$ $H e i s r i g h t$
	Answered by Hassen_Timol last updated on 12/Mar/19

	$It is when x \to - \infty or x \to + \infty I t h i n k . . .$
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