Solve-for-x-R-x-2-x-1-x-2-x-1-x-3-1-x-3-1-

Question Number 148689 by liberty last updated on 30/Jul/21

Solve for x \in R

\frac{x^{2} - x + 1}{x^{2} + x + 1} = \sqrt{\frac{x^{3} - 1}{x^{3} + 1}} .

Answered by MJS_new last updated on 30/Jul/21

squaring (beware of false solutions!) and

transforming leads to

x^{6} + \frac{1}{2} x^{4} - \frac{3}{2} x^{2} - \frac{1}{2} = 0

let x = \pm \sqrt{z - \frac{1}{6}}

z^{3} - \frac{19}{12} z - \frac{13}{54} = 0

z_{1} = - \frac{\sqrt{19}}{3} \cos (\frac{π}{6} + \frac{1}{3} \arcsin \frac{26 \sqrt{19}}{361}) \approx - 1.17399865514

z_{2} = - \frac{\sqrt{19}}{3} \sin (\frac{1}{3} \arcsin \frac{26 \sqrt{19}}{361}) \approx - . 154370149574

z_{3} = \frac{\sqrt{19}}{3} \sin (\frac{π}{3} + \frac{1}{3} \arcsin \frac{26 \sqrt{19}}{361}) \approx 1.32836880471

\Rightarrow

x_{1} \approx \pm 1 . 15787102987 i

x_{2} \approx \pm . 566601108577 i

x_{3} \approx \pm 1.07782286951

testing shows only x_{2} and x_{3} solve the given

equation . since x \in R we get

x \approx \pm 1.07782286951

Commented by liberty last updated on 30/Jul/21

Commented by liberty last updated on 30/Jul/21

correct sir . thanks

Answered by Rasheed.Sindhi last updated on 30/Jul/21

\frac{x^{2} - x + 1}{x^{2} + x + 1} = \sqrt{\frac{x^{3} - 1}{x^{3} + 1}}

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

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

{(\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}})}^{2} - \sqrt{\frac{x - 1}{x + 1}} . {(\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}})}^{- 1} = 0

\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}} (\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}} - \sqrt{\frac{x - 1}{x + 1}} . {(\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}})}^{- 2}) = 0

\sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}} = 0 ∣ \sqrt{\frac{x^{2} - x + 1}{x^{2} + x + 1}} = \sqrt{\frac{x - 1}{x + 1}} {(\sqrt{\frac{x^{2} + x + 1}{x^{2} - x + 1}})}^{2}

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

(x + 1) {(x^{2} - x + 1)}^{3} = (x - 1) {(x^{2} + x + 1)}^{3}

(x^{3} + 1) {(x^{2} - x + 1)}^{2} = (x^{3} - 1) {(x^{2} + x + 1)}^{2}

\dots . .

\dots .

Commented by Rasheed.Sindhi last updated on 30/Jul/21

S i r M J S, p l e a s e g u i d e m e w h e r e

I l o s e r e a l r o o t s ?

Commented by MJS_new last updated on 30/Jul/21

your mistake is between lines 4 and 5

line 4 : {(\sqrt{a})}^{2} - \sqrt{b} \sqrt{a^{- 1}} = 0

\overset{? ? ?}{\Rightarrow}

line 5 : \sqrt{a^{- 1}} (\sqrt{a^{- 1}} - \sqrt{b}) = 0

Commented by Rasheed.Sindhi last updated on 30/Jul/21

Y e s s i r, T h a n X a L o t!

Commented by MJS_new last updated on 30/Jul/21

{you}^{'} re welcome as always!

Commented by Rasheed.Sindhi last updated on 31/Jul/21

S i r, o n e t h i n g I f e e l t h a t n o w a d a y s

y o u {a r e n}^{'} t c o n t r i b u t i n g v e r y m u c h ?

Commented by ajfour last updated on 31/Jul/21

i h a d n o t a n i d e a, i t h a d t o d o

a n y t h i n g w i t h t h e R i e m a n n

t h i n g, t h a t q u e s t i o n j u s t s u r f a c e d

o n m y m i n d . t h a n k s f o r t h e

l i n k . i h a v e b o o k m a r k e d i t,

i n d e f i n i t e p o s t p o n e m e n t .

Commented by MJS_new last updated on 31/Jul/21

to Sir Rasheed Sindhi : I feel the quality of

questions has been dropping . I stopped

answering certain people because I {don}^{'} t

do their homework for free and on the other

hand there are those who ask for the solution

of {R i e m a n n}^{'} s C o n j e c t u r e on one day and

for the result of simple fractional arithmetics

the other day \dots

Commented by ajfour last updated on 31/Jul/21

t h e s s y m p t o t e t h i n g w e a r e

s u r e i c a n d o m y s e l f, a g a i n s o r r y

t o h a v e a s k e d y o u t h a t .

Commented by MJS_new last updated on 31/Jul/21

I {didn}^{'} t mean to upset you, as you are one

of the few persons who are the reason I^{'} m

still here . asymptotes of what ? I guess I

missed something, sorry

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