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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∈R    ((x^2 −x+1)/(x^2 +x+1)) = (√((x^3 −1)/(x^3 +1))) .
$$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\in\mathrm{R} \\ $$$$\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}\:. \\ $$
Answered by MJS_new last updated on 30/Jul/21
squaring (beware of false solutions!) and  transforming leads to  x^6 +(1/2)x^4 −(3/2)x^2 −(1/2)=0  let x=±(√(z−(1/6)))  z^3 −((19)/(12))z−((13)/(54))=0  z_1 =−((√(19))/3)cos ((π/6)+(1/3)arcsin ((26(√(19)))/(361))) ≈−1.17399865514  z_2 =−((√(19))/3)sin ((1/3)arcsin ((26(√(19)))/(361))) ≈−.154370149574  z_3 =((√(19))/3)sin ((π/3)+(1/3)arcsin ((26(√(19)))/(361))) ≈1.32836880471  ⇒  x_1 ≈±1.15787102987i  x_2 ≈±.566601108577i  x_3 ≈±1.07782286951  testing shows only x_2  and x_3  solve the given  equation. since x∈R we get  x≈±1.07782286951
$$\mathrm{squaring}\:\left(\mathrm{beware}\:\mathrm{of}\:\mathrm{false}\:\mathrm{solutions}!\right)\:\mathrm{and} \\ $$$$\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}=\pm\sqrt{{z}−\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{19}}{\mathrm{12}}{z}−\frac{\mathrm{13}}{\mathrm{54}}=\mathrm{0} \\ $$$${z}_{\mathrm{1}} =−\frac{\sqrt{\mathrm{19}}}{\mathrm{3}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{26}\sqrt{\mathrm{19}}}{\mathrm{361}}\right)\:\approx−\mathrm{1}.\mathrm{17399865514} \\ $$$${z}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{19}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{26}\sqrt{\mathrm{19}}}{\mathrm{361}}\right)\:\approx−.\mathrm{154370149574} \\ $$$${z}_{\mathrm{3}} =\frac{\sqrt{\mathrm{19}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{26}\sqrt{\mathrm{19}}}{\mathrm{361}}\right)\:\approx\mathrm{1}.\mathrm{32836880471} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} \approx\pm\mathrm{1}.\mathrm{15787102987i} \\ $$$${x}_{\mathrm{2}} \approx\pm.\mathrm{566601108577i} \\ $$$${x}_{\mathrm{3}} \approx\pm\mathrm{1}.\mathrm{07782286951} \\ $$$$\mathrm{testing}\:\mathrm{shows}\:\mathrm{only}\:{x}_{\mathrm{2}} \:\mathrm{and}\:{x}_{\mathrm{3}} \:\mathrm{solve}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation}.\:\mathrm{since}\:{x}\in\mathbb{R}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}\approx\pm\mathrm{1}.\mathrm{07782286951} \\ $$
Commented by liberty last updated on 30/Jul/21
Commented by liberty last updated on 30/Jul/21
correct sir. thanks
$$\mathrm{correct}\:\mathrm{sir}.\:\mathrm{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Jul/21
  ((x^2 −x+1)/(x^2 +x+1)) = (√((x^3 −1)/(x^3 +1)))    ((√((x^2 −x+1)/(x^2 +x+1))))^2  = (√(((x−1)(x^2 +x+1))/((x+1)(x^2 −x+1))))   ((√((x^2 −x+1)/(x^2 +x+1))))^2 = (√((x−1)/(x+1))).(√((x^2 +x+1)/(x^2 −x+1)))   ((√((x^2 −x+1)/(x^2 +x+1))))^2 − (√((x−1)/(x+1))).((√((x^2 −x+1)/(x^2 +x+1))))^(−1) =0  (√((x^2 −x+1)/(x^2 +x+1)))((√((x^2 −x+1)/(x^2 +x+1)))−(√((x−1)/(x+1))).((√((x^2 −x+1)/(x^2 +x+1))))^(−2) )=0   (√((x^2 −x+1)/(x^2 +x+1)))=0 ∣ (√((x^2 −x+1)/(x^2 +x+1)))=(√((x−1)/(x+1)))((√((x^2 +x+1)/(x^2 −x+1))))^2    ((x^2 −x+1)/(x^2 +x+1))=((x−1)/(x+1))(((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   .....  ....
$$\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}} \\ $$$$\:\:\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\right)^{\mathrm{2}} \:=\:\sqrt{\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)}} \\ $$$$\:\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\right)^{\mathrm{2}} =\:\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}}.\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}} \\ $$$$\:\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\right)^{\mathrm{2}} −\:\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}}.\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\right)^{−\mathrm{1}} =\mathrm{0} \\ $$$$\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}−\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}}.\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\right)^{−\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}=\mathrm{0}\:\mid\:\sqrt{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}=\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}}\left(\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}}\right)^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}=\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} =\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$….. \\ $$$$…. \\ $$
Commented by Rasheed.Sindhi last updated on 30/Jul/21
Sir MJS, please guide me where  I lose real roots?
$${Sir}\:{MJS},\:{please}\:{guide}\:{me}\:{where} \\ $$$${I}\:{lose}\:{real}\:{roots}? \\ $$
Commented by MJS_new last updated on 30/Jul/21
your mistake is between lines 4 and 5  line 4: ((√a))^2 −(√b)(√a^(−1) )=0  ⇒^(???)   line 5: (√a^(−1) )((√a^(−1) )−(√b))=0
$$\mathrm{your}\:\mathrm{mistake}\:\mathrm{is}\:\mathrm{between}\:\mathrm{lines}\:\mathrm{4}\:\mathrm{and}\:\mathrm{5} \\ $$$$\mathrm{line}\:\mathrm{4}:\:\left(\sqrt{{a}}\right)^{\mathrm{2}} −\sqrt{{b}}\sqrt{{a}^{−\mathrm{1}} }=\mathrm{0} \\ $$$$\overset{???} {\Rightarrow} \\ $$$$\mathrm{line}\:\mathrm{5}:\:\sqrt{{a}^{−\mathrm{1}} }\left(\sqrt{{a}^{−\mathrm{1}} }−\sqrt{{b}}\right)=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 30/Jul/21
Yes sir, ThanX a Lot!
$${Yes}\:\boldsymbol{{sir}},\:\boldsymbol{\mathcal{T}}{han}\mathcal{X}\:{a}\:\boldsymbol{\mathcal{L}}{ot}! \\ $$
Commented by MJS_new last updated on 30/Jul/21
you′re welcome as always!
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{as}\:\mathrm{always}! \\ $$
Commented by Rasheed.Sindhi last updated on 31/Jul/21
Sir, one thing I feel that nowadays  you aren′t contributing very much?
$${Sir},\:{one}\:{thing}\:{I}\:{feel}\:{that}\:{nowadays} \\ $$$${you}\:{aren}'{t}\:{contributing}\:{very}\:{much}? \\ $$
Commented by ajfour last updated on 31/Jul/21
i had not an idea, it had to do  anything with the Riemann  thing, that question just surfaced  on my mind. thanks for the   link. i have bookmarked it,  indefinite postponement.
$${i}\:{had}\:{not}\:{an}\:{idea},\:{it}\:{had}\:{to}\:{do} \\ $$$${anything}\:{with}\:{the}\:{Riemann} \\ $$$${thing},\:{that}\:{question}\:{just}\:{surfaced} \\ $$$${on}\:{my}\:{mind}.\:{thanks}\:{for}\:{the}\: \\ $$$${link}.\:{i}\:{have}\:{bookmarked}\:{it}, \\ $$$${indefinite}\:{postponement}. \\ $$
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 Riemann′s Conjecture on one day and  for the result of simple fractional arithmetics  the other day...
$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Rasheed}\:\mathrm{Sindhi}:\:\mathrm{I}\:\mathrm{feel}\:\mathrm{the}\:\mathrm{quality}\:\mathrm{of} \\ $$$$\mathrm{questions}\:\mathrm{has}\:\mathrm{been}\:\mathrm{dropping}.\:\mathrm{I}\:\mathrm{stopped} \\ $$$$\mathrm{answering}\:\mathrm{certain}\:\mathrm{people}\:\mathrm{because}\:\mathrm{I}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{do}\:\mathrm{their}\:\mathrm{homework}\:\mathrm{for}\:\mathrm{free}\:\mathrm{and}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{hand}\:\mathrm{there}\:\mathrm{are}\:\mathrm{those}\:\mathrm{who}\:\mathrm{ask}\:\mathrm{for}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{of}\:{Riemann}'{s}\:{Conjecture}\:\mathrm{on}\:\mathrm{one}\:\mathrm{day}\:\mathrm{and} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{result}\:\mathrm{of}\:\mathrm{simple}\:\mathrm{fractional}\:\mathrm{arithmetics} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{day}… \\ $$
Commented by ajfour last updated on 31/Jul/21
the ssymptote thing we are  sure i can do myself, again sorry  to have asked you that.
$${the}\:{ssymptote}\:{thing}\:{we}\:{are} \\ $$$${sure}\:{i}\:{can}\:{do}\:{myself},\:{again}\:{sorry} \\ $$$${to}\:{have}\:{asked}\:{you}\:{that}. \\ $$
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
$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{mean}\:\mathrm{to}\:\mathrm{upset}\:\mathrm{you},\:\mathrm{as}\:\mathrm{you}\:\mathrm{are}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{few}\:\mathrm{persons}\:\mathrm{who}\:\mathrm{are}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{I}'\mathrm{m} \\ $$$$\mathrm{still}\:\mathrm{here}.\:\mathrm{asymptotes}\:\mathrm{of}\:\mathrm{what}?\:\mathrm{I}\:\mathrm{guess}\:\mathrm{I} \\ $$$$\mathrm{missed}\:\mathrm{something},\:\mathrm{sorry} \\ $$

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