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

Question Number 53262 by Tawa1 last updated on 19/Jan/19

find x : \frac{1}{\sqrt{x + 1 + \sqrt{x}}} - \frac{2}{\sqrt{x - 2 + \sqrt{x}}} = \sqrt{x - 1}

Commented by mr W last updated on 20/Jan/19

t h e r e i s n o r e a l s o l u t i o n!

L H S :

s a y x + \sqrt{x} = u > 2

\sqrt{u + 1} > \sqrt{u - 2} > 0

\Rightarrow \frac{1}{\sqrt{u + 1}} < \frac{1}{\sqrt{u - 2}} < \frac{2}{\sqrt{u - 2}}

\Rightarrow L H S = \frac{1}{\sqrt{x + 1 + \sqrt{x}}} - \frac{2}{\sqrt{x - 2 + \sqrt{x}}} = \frac{1}{\sqrt{u + 1}} - \frac{2}{\sqrt{u - 2}} < 0

R H S :

\sqrt{x - 1} ⩾ 0

i . e . L H S i s a l w a y s n e g a t i v e, b u t R H S

i s a l w a y s p o s i t i v e o r z e r o .

L H S < 0 \neq R H S ⩾ 0

\Rightarrow n o s o l u t i o n f o r L H S = R H S!

Commented by Tawa1 last updated on 20/Jan/19

God bless you sir

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

c o n d i t i o n \dots

1) x > 0

2) (x - 1) > 0 i, e x > 1

3) x - 2 + \sqrt{x} > 0

i f x = 1 t h e n x - 2 + \sqrt{x} = 0

b u t v a l u e o f x + \sqrt{x} s h o u l d b e g r e a t e r t h a n 2

s o x s h o u l d b e x > 1

4) x + 1 + \sqrt{x} > 0 w h e n x > 1

s o x s h o u l d b e g r e a t e r t h a n 1

w a i t \dots

f (x) = \frac{1}{\sqrt{x + \sqrt{x} + + 1}}

g (x) = \sqrt{x - 1} + \frac{1}{\sqrt{x + \sqrt{x} - 2}}

f r o m g r a p h i t i s c o n f i r m e d t h a t f (x) n e v e r

i n t e r s e c t g (x) h e n c e n o s o l u t i o n \dots

h o w e v e r \dots

c o n d i t i o n o f t w o c u r v e i n t e s e c t / t o u c h i s s h o r t e s t

d i s t a n c e z e r o . .

Commented by Tawa1 last updated on 20/Jan/19

Alright sir . I wait . Thanks for your time sir .

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19