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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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