solve for x∈R (√(x−(1/x)))+(√(1−(1/x)))=x


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Question Number 200498 by mr W last updated on 19/Nov/23

$s o l v e f o r x \in R$ $\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
Commented by BaliramKumar last updated on 19/Nov/23

$Q . 192958$
Commented by mr W last updated on 19/Nov/23

$r e p o s t e d f o r a l t e r n a t i v e s o l u t i o n s .$
Commented by Harnada last updated on 20/Nov/23

$Q 183839$
	Answered by cortano12 last updated on 19/Nov/23

	$\underset{―}{\underset{⏟}{}}$
	Answered by Rasheed.Sindhi last updated on 19/Nov/23

	$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x . . . . . . . . . (i)$ $(x - \frac{1}{x}) - (1 - \frac{1}{x}) = x (\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}})$ $\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = \frac{x - 1}{x} . . . (i i)$ $(i) + (i i) :$ $2 \sqrt{x - \frac{1}{x}} = x - \frac{1}{x} + 1$ $\sqrt{x - \frac{1}{x}} = y (s a y)$ $2 y = y^{2} + 1$ $y^{2} - 2 y + 1 = 0$ ${(y - 1)}^{2} = 0$ $y = 1$ $x - \frac{1}{x} = 1$ $x^{2} - x - 1 = 0$ $x_{⩾ 0} = \frac{1 + \sqrt{1 + 4}}{2} = \frac{1 + \sqrt{5}}{2}$
	Answered by Rasheed.Sindhi last updated on 19/Nov/23

	$a + b = x . . . . . . . . . (i)$ $a^{2} - b^{2} = (x - \frac{1}{x}) - (1 - \frac{1}{x}) = x - 1$ $a - b = \frac{a^{2} - b^{2}}{a + b} = \frac{x - 1}{x} . . . . . . (i i)$ $(i) + (i i) :$ $2 a = x + \frac{x - 1}{x} = x - \frac{1}{x} + 1$ $2 (\underset{y}{\underset{⏟}{\sqrt{x - \frac{1}{x}}}}) = x - \frac{1}{x} + 1$ $2 y = y^{2} + 1$ ${(y - 1)}^{2} = 0$ $y = 1$ $\sqrt{x - \frac{1}{x}} = 1$ $x - \frac{1}{x} = 1$ $x^{2} - x - 1 = 0$ $x_{⩾ 0} = \frac{1 + \sqrt{5}}{2}$
	Answered by Frix last updated on 19/Nov/23

	$\sqrt{x - \frac{1}{x}} = u \land \sqrt{1 - \frac{1}{x}} = v \Rightarrow x = u + v; u, v ⩾ 0$ $(1) u + v - \frac{1}{u + v} = u^{2} \Rightarrow u^{3} + u^{2} (v - 1) - 2 u v - v^{2} + 1 = 0$ $(2) 1 - \frac{1}{u + v} = v^{2} \Rightarrow u = \frac{v^{3} - v + 1}{1 - v^{2}}$ $(1) \frac{{(v^{2} + v - 1)}^{2}}{{(1 - v^{2})}^{3}} = 0 \Rightarrow v = - \frac{1}{2} + \frac{\sqrt{5}}{2} \Rightarrow u = 1$ $x = u + v = \frac{1}{2} + \frac{\sqrt{5}}{2}$
	Answered by mr W last updated on 19/Nov/23

	Commented by Rasheed.Sindhi last updated on 20/Nov/23

	$G reat!$ $T h a n k s s i r!$
	Commented by mr W last updated on 19/Nov/23

	$x > 0$ ${(\sqrt{x})}^{2} + 1^{2} = x^{2}$ $x^{2} - x - 1 = 0$ $\Rightarrow x = \frac{1 + \sqrt{5}}{2}$
	Commented by Rasheed.Sindhi last updated on 20/Nov/23

	$Y o u s e e g e o m e t r y i n a l g e b r a$ $w h e r e o t h e r s d o n o t s e e!$
	Commented by Rasheed.Sindhi last updated on 20/Nov/23

	$B u t s i r I {c a n}^{'} t u n d e r s t a n d h o w i s$ $t h e a n s w e r r e l a t e d t o t h e q u e s t i o n ?$
	Commented by manolex last updated on 20/Nov/23

	$i n g r e d i e n t s : \sqrt{x - \frac{1}{x}}, \sqrt{1 - \frac{1}{x}}, x p r o d u c t t r i a n g l e 1, x, \sqrt{x}, h = \frac{1}{\sqrt{x}} w a y s u s t e n t a t i o n . s a l u d o s m r W$
	Commented by mr W last updated on 20/Nov/23

	$t h a n k s y o u s i r!$ $t o b e h o n e s t, i h a v e a f o n d n e s s f o r$ $g e o m e t r y . a t f i r s t i a l w a y s t r y t o$ $u n d e r s t a n d i f t h e r e i s a g e o m e t r i c a l$ $m e a n i n g o r i n t e r p r e t a t i o n f o r t h e$ $q u e s t i o n . i n t h i s w a y i h a v e s o l v e d$ $s o m e q u e s t i o n s i n t h i s f o r u m w h i c h$ $a t t h e f i r s t g l a n c e h a v e n o t h i n g t o d o$ $w i t h g e o m e t r y .$
	Commented by mr W last updated on 20/Nov/23

	Commented by mr W last updated on 20/Nov/23

	$B D = \sqrt{{(\sqrt{x})}^{2} - {(\frac{1}{\sqrt{x}})}^{2}} = \sqrt{x - \frac{1}{x}}$ $D C = \sqrt{1^{2} - {(\frac{1}{\sqrt{x}})}^{2}} = \sqrt{1 - \frac{1}{x}}$ $B C = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x (a s g i v e n)$ $\frac{A B}{A C} = \frac{\sqrt{x}}{1} = \sqrt{x} = \frac{B C}{A B} = \frac{x}{\sqrt{x}} = \sqrt{x} = \frac{A C}{A D} = \frac{1}{\frac{1}{\sqrt{x}}} = \sqrt{x}$ $\Rightarrow Δ A B C i s r i g h t - a n g l e d t r i a n g l e$ $\Rightarrow {(\sqrt{x})}^{2} + 1^{2} = x^{2}$ $\Rightarrow x^{2} - x - 1 = 0$
	Commented by necx122 last updated on 20/Nov/23
	I've always suspected the same that you and Ajfour had a special likeness for geometry because your manner of approach tackling geometry based questions is most often overwhelming.
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