solve-for-x-R-x-1-x-1-1-x-x-

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

(√(x−(1/x)))+(√(1−(1/x))) =x.........(i) (x−(1/x))−(1−(1/x))=x((√(x−(1/x))) −(√(1−(1/x))) ) (√(x−(1/x))) −(√(1−(1/x))) =((x−1)/x)...(ii) (i)+(ii): 2(√(x−(1/x))) =x−(1/x)+1 (√(x−(1/x))) =y (say) 2y=y^2 +1 y^2 −2y+1=0 (y−1)^2 =0 y=1 x−(1/x)=1 x^2 −x−1=0 x_(≥0) =((1+(√(1+4)))/2)=((1+(√5))/2)

\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x \dots \dots \dots (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} \dots (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−(1/x))−(1−(1/x))=x−1 a−b=((a^2 −b^2 )/(a+b))=((x−1)/x)......(ii) (i)+(ii): 2a=x+((x−1)/x)=x−(1/x)+1 2((√(x−(1/x)))_(y) )=x−(1/x)+1 2y=y^2 +1 (y−1)^2 =0 y=1 (√(x−(1/x))) =1 x−(1/x)=1 x^2 −x−1=0 x_(≥0) =((1+(√5) )/2)

a + b = x \dots \dots \dots (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} \dots \dots (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

(√(x−(1/x)))=u∧(√(1−(1/x)))=v ⇒ x=u+v; u, v ≥0 (1) u+v−(1/(u+v))=u^2 ⇒ u^3 +u^2 (v−1)−2uv−v^2 +1=0 (2) 1−(1/(u+v))=v^2 ⇒ u=((v^3 −v+1)/(1−v^2 )) (1) (((v^2 +v−1)^2 )/((1−v^2 )^3 ))=0 ⇒ v=−(1/2)+((√5)/2) ⇒ u=1 x=u+v=(1/2)+((√5)/2)

\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 ((√x))^2 +1^2 =x^2 x^2 −x−1=0 ⇒x=((1+(√5))/2)

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

You see geometry in algebra where others do not see!

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

But sir I can′t understand how is the answer related to the question?

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

ingredients: (√(x−(1/x) )),(√(1−(1/x))) ,x product triangle 1,x,(√x),h=(1/( (√x))) way sustentation .saludos mr W

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

thanks you sir! to be honest, i have a fondness for geometry. at first i always try to understand if there is a geometrical meaning or interpretation for the question. in this way i have solved some questions in this forum which at the first glance have nothing to do with geometry.

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

BD=(√(((√x))^2 −((1/( (√x))))^2 ))=(√(x−(1/x))) DC=(√(1^2 −((1/( (√x))))^2 ))=(√(1−(1/x))) BC=(√(x−(1/x)))+(√(1−(1/x)))=x (as given) ((AB)/(AC))=((√x)/1)=(√x)=((BC)/(AB))=(x/( (√x)))=(√x)=((AC)/(AD))=(1/(1/( (√x))))=(√x) ⇒ΔABC is right−angled triangle ⇒((√x))^2 +1^2 =x^2 ⇒x^2 −x−1=0

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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