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Question Number 182852 by mnjuly1970 last updated on 16/Dec/22

$$$$$$challangingquestion..$$$$if,\frac{1}{1+\sqrt{1-x+x^2}}-\frac{x}{1+\sqrt{1+x+x^2}}=1$$$$findthevalueof:\frac{x}{x^2+1}$$$$$$

Commented by Frix last updated on 15/Dec/22

$$\mathrm{The}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{are}$$$$x=\frac{6+\sqrt{-34+10\sqrt{13}}}{10}\pm \frac{2+\sqrt{34+10\sqrt{13}}}{10}\mathrm{i}$$$$\Rightarrow \frac{x}{x^2+1}=\frac{3\pm \mathrm{i}}{4}$$

Commented by Frix last updated on 16/Dec/22

$$\mathrm{Now}\mathrm{you}\mathrm{changed}\mathrm{it}.$$$$\mathrm{Now}x=-\frac{1+\sqrt{7}+\sqrt{-1+2\sqrt{7}}}{3}\Rightarrow$$$$\frac{x}{x^2+1}=\frac{1-\sqrt{7}}{4}$$

Answered by Rasheed.Sindhi last updated on 16/Dec/22

$$\frac{1}{1+\sqrt{1-x+x^2}}+\frac{x}{1+\sqrt{1+x+x^2}}=1$$$$\frac{x}{x^2+1}=?$$$$\frac{1}{1+\sqrt{x(x+\frac{1}{x}-1)}}+\frac{x}{1+\sqrt{x(x+\frac{1}{x}+1)}}=1$$$$\frac{1}{1+\sqrt{x}\sqrt{x+\frac{1}{x}-1}}+\frac{x}{1+\sqrt{x}\sqrt{x+\frac{1}{x}+1}}=1$$$$Letx+\frac{1}{x}+1=y^2$$$$x+\frac{1}{x}-1=y^2-2$$$$\frac{1}{1+\sqrt{x}\sqrt{y^2-2}}+\frac{x}{1+y\sqrt{x}}=1$$$$1-\frac{1}{1+\sqrt{x}\sqrt{y^2-2}}=\frac{x}{1+y\sqrt{x}}$$$$\frac{1+\sqrt{x}\sqrt{y^2-2}-1}{1+\sqrt{x}\sqrt{y^2-2}}=\frac{x}{1+y\sqrt{x}}$$$$\frac{\sqrt{x}\sqrt{y^2-2}}{1+\sqrt{x}\sqrt{y^2-2}}=\frac{x}{1+y\sqrt{x}}$$$$\frac{1+\sqrt{x}\sqrt{y^2-2}}{\sqrt{x}\sqrt{y^2-2}}=\frac{1+y\sqrt{x}}{x}$$$$\frac{1}{\sqrt{x}\sqrt{y^2-2}}+1=\frac{1}{x}+\frac{y}{\sqrt{x}}$$$$\frac{1}{\sqrt{x}\sqrt{y^2-2}}-\frac{y}{\sqrt{x}}=\frac{1}{x}-1=\frac{1-x}{x}$$$$\frac{1}{\sqrt{y^2-2}}-y=\frac{1-x}{\sqrt{x}}$$$$\frac{1-y\sqrt{y^2-2}}{\sqrt{y^2-2}}=\frac{1-x}{\sqrt{x}}$$$${\left(\frac{1-y\sqrt{y^2-2}}{\sqrt{y^2-2}}\right)}^2={\left(\frac{1-x}{\sqrt{x}}\right)}^2$$$$\frac{1+y^2(y^2-2)-2y\sqrt{y^2-2}}{y^2-2}=\frac{1+x^2-2x}{x}$$$$\frac{1+y^2(y^2-2)-2y\sqrt{y^2-2}}{y^2-2}=x+\frac{1}{x}-2$$$$\frac{1+y^2(y^2-2)-2y\sqrt{y^2-2}}{y^2-2}=x+\frac{1}{x}+1-3$$$$\frac{1+y^2(y^2-2)-2y\sqrt{y^2-2}}{y^2-2}=y^2-3$$$$1+y^4-2y^2-2y\sqrt{y^2-2}=(y^2-3)(y^2-2)$$$$1+y^4-2y^2-2y\sqrt{y^2-2}=y^4-5y^2+6$$$$2y\sqrt{y^2-2}=3y^2-5$$$$4y^2(y^2-2)=9y^4-30y^2+25$$$$5y^4-22y^2+25=0$$$$y^2=\frac{22\pm \sqrt{484-500}}{10}=\frac{22\pm 4\mathit{i}}{10}=\frac{11\pm 2\mathit{i}}{5}$$$$$$$$▸\frac{x}{x^2+1}=\frac{1}{x+\frac{1}{x}}=\frac{1}{x+\frac{1}{x}+1-1}=\frac{1}{y^2-1}$$$$=\frac{1}{\frac{11\pm 2\mathit{i}}{5}-1}=\frac{1}{\frac{6\pm 2i}{5}}=\frac{5}{6\pm 2i}\cdot \frac{6\mp 2i}{6\mp 2i}$$$$=\frac{5(6\mp 2i)}{36+4}=\frac{6\mp 2i}{8}=\frac{3\pm i}{4}$$

Commented by manxsol last updated on 16/Dec/22

$$Iamlookingfora$$$$simpler$$$$development\left(desarrollo\right)$$$$.Greetings$$

Commented by manxsol last updated on 16/Dec/22

$$errorinline$$$$2y\sqrt{y^2-2}=3y^2-5$$$$4y^2(y^2-2)=3y^2-5$$$$4y^2(y^2-2)=9y^4-30y^2+25$$$$5y^4-22y^2+25=0$$$$y^2=\frac{22\pm \sqrt{{22}^2-500}}{10}$$$$y^2=\frac{22\pm 4\mathit{i}}{10}$$$$y^2=\frac{11\pm 2\mathit{i}}{5}$$$$\frac{x}{x^2+1}=?\Rightarrow \frac{1}{x+\frac{1}{x}}=\frac{1}{y^2-1}$$$$\frac{1}{\frac{11\pm 2i}{5}-1}=\frac{1}{\frac{6\pm 2i}{5}}=\frac{5}{6\pm 2i}=\frac{5(6-2i)}{40}$$$$\frac{x}{x^2+1}=\frac{3\pm i}{4}$$$$$$

Commented by Rasheed.Sindhi last updated on 15/Dec/22

$$\mathit{P}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}checkmyanswerforerrors.$$

Commented by mnjuly1970 last updated on 15/Dec/22

$$x\in \mathbb{R}\Rightarrow putx=tan\left(\alpha \right)andsolve$$$$answeris\frac{1-\sqrt{7}}{4}$$

Commented by mr W last updated on 15/Dec/22

$$thensomethingiswronginquestion.$$$$forx\in R,ithink$$$$\frac{1}{1+\sqrt{1-x+x^2}}+\frac{x}{1+\sqrt{1+x+x^2}}=1has$$$$nosolution.$$

Commented by ellenpaulisrae last updated on 15/Dec/22

$$hello$$

Commented by Frix last updated on 15/Dec/22

$$\mathrm{Yes},{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{real}\mathrm{solution}\mathrm{for}\mathrm{the}\mathrm{given}$$$$\mathrm{equation}.$$

Commented by Rasheed.Sindhi last updated on 16/Dec/22

$$\mathcal{T}han\mathcal{X}manxsol\mathit{s}\mathit{i}\mathit{r}!$$$$I^{\prime }mgoingtocorrectmyanswer.$$$$\mathcal{T}heapproachiscorrectalthough$$$$lengthy!I^{\prime }llwaitforyoursimpler$$$$approach.Butmeanwhilemnjulysir$$$$haschangedhisquestionandvery$$$$sorrytosaythathedidthiswithout$$$$any\mathit{i}\mathit{n}\mathit{t}\mathit{i}\mathit{m}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}!Anyway\mathrm{than}\mathbb{k}\mathrm{s}$$$$\mathrm{again}!$$

Commented by mnjuly1970 last updated on 16/Dec/22

$$youarerightsirW$$$$imodifidthequestion..$$

Answered by Rasheed.Sindhi last updated on 16/Dec/22

$$\mathrm{Changed}\mathrm{version}\mathrm{of}\mathrm{the}\mathrm{question}:$$$$\frac{1}{1+\sqrt{1-x+x^2}}-\frac{x}{1+\sqrt{1+x+x^2}}=1$$$$\frac{1}{1+\sqrt{x}\left(\sqrt{x+\frac{1}{x}-1}\right)}-1=\frac{x}{1+\sqrt{x}\left(\sqrt{x+\frac{1}{x}+1}\right)}$$$$Letx+\frac{1}{x}+1=y^2\Rightarrow x+\frac{1}{x}-1=y^2-2$$$$\frac{1-1-\sqrt{x}\left(\sqrt{y^2-2}\right)}{1+\sqrt{x}\left(\sqrt{y^2-2}\right)}=\frac{x}{1+y\sqrt{x}}$$$$\frac{-\sqrt{x}\left(\sqrt{y^2-2}\right)}{1+\sqrt{x}\left(\sqrt{y^2-2}\right)}=\frac{x}{1+y\sqrt{x}}$$$$\frac{-1-\sqrt{x}\left(\sqrt{y^2-2}\right)}{\sqrt{x}\left(\sqrt{y^2-2}\right)}=\frac{1+y\sqrt{x}}{x}$$$$\frac{-1}{\sqrt{x}(\sqrt{y^2-2}}-1=\frac{1}{x}+\frac{y}{\sqrt{x}}$$$$\frac{y}{\sqrt{x}}+\frac{1}{\sqrt{x}\left(\sqrt{y^2-2}\right)}=-1-\frac{1}{x}=-\frac{x+1}{x}$$$$\frac{y}{\sqrt{x}}+\frac{1}{\sqrt{x}\left(\sqrt{y^2-2}\right)}=-\left(\frac{x+1}{x}\right)$$$$\frac{y\sqrt{y^2-2}+1}{\sqrt{y^2-2}}=-\frac{x+1}{\sqrt{x}}$$$${\left(\frac{y\sqrt{y^2-2}+1}{\sqrt{y^2-2}}\right)}^2={(-\frac{x+1}{\sqrt{x}})}^2$$$$\frac{y^2(y^2-2)+1+2y\sqrt{y^2-2}}{y^2-2}=\frac{x^2+2x+1}{x}$$$$\frac{y^4-2y^2+1+2y\sqrt{y^2-2}}{y^2-2}=x+\frac{1}{x}+2=x+\frac{1}{x}+1+1$$$$y^4-2y^2+1+2y\sqrt{y^2-2}=(y^2+1)(y^2-2)$$$$y^4-2y^2+1+2y\sqrt{y^2-2}=y^4-y^2-2$$$$2y\sqrt{y^2-2}=y^2-3$$$$4y^2(y^2-2)=y^4-6y^2+9$$$$4y^4-8y^2-y^4+6y^2-9=0$$$$3y^4-2y^2-9=0$$$$y^2=\frac{2\pm \sqrt{4+108}}{6}=\frac{2\pm \sqrt{112}}{6}=\frac{1\pm 2\sqrt{7}}{3}$$$$\because y^{2}isnon-negative$$$$\therefore y^2=\frac{1+2\sqrt{7}}{3}$$$$\frac{x}{x^2+1}=\frac{1}{x+\frac{1}{x}+1-1}=\frac{1}{y^2-1}$$$$=\frac{1}{\frac{1+2\sqrt{7}}{3}-1}=\frac{3}{-2+2\sqrt{7}}\cdot \frac{-2-2\sqrt{7}}{-2-2\sqrt{7}}$$$$\frac{3(-2+2\sqrt{7})}{4-28}y=\frac{6(-1+\sqrt{7})}{-24}=\frac{1-\sqrt{7}}{4}$$

Commented by mnjuly1970 last updated on 16/Dec/22

$$verynicesolution$$$$sirRasheed.thanksalot$$$$foryoureffort.$$$$thatwasmymistake⟨typo⟩$$$$imodifiedit.$$

Answered by Rasheed.Sindhi last updated on 17/Dec/22

$$\mathit{S}\mathit{i}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathit{r}\mathit{w}\mathit{a}\mathit{y}$$$$\frac{1}{1+\sqrt{1-x+x^2}}-\frac{x}{1+\sqrt{1+x+x^2}}=1$$$$\frac{x}{x^2+1}=?$$$$\frac{x}{x^2+1}=\frac{1}{y}\Rightarrow x^2+1=xy$$$$\frac{1}{1+\sqrt{xy-x}}-\frac{x}{1+\sqrt{xy+x}}=1$$$$\frac{1}{1+\sqrt{x}\sqrt{y-1}}-\frac{x}{1+\sqrt{x}\sqrt{y+1}}=1$$$$\frac{1}{1+\sqrt{x}\sqrt{y-1}}-1=\frac{x}{1+\sqrt{x}\sqrt{y+1}}$$$$\frac{-\sqrt{x}\sqrt{y-1}}{1+\sqrt{x}\sqrt{y-1}}=\frac{x}{1+\sqrt{x}\sqrt{y+1}}$$$$\frac{-1-\sqrt{x}\sqrt{y-1}}{\sqrt{x}\sqrt{y-1}}=\frac{1+\sqrt{x}\sqrt{y+1}}{x}$$$$\frac{-1}{\sqrt{x}\sqrt{y-1}}-1=\frac{1}{x}+\frac{\sqrt{y+1}}{\sqrt{x}}$$$$\frac{\sqrt{y+1}}{\sqrt{x}}+\frac{1}{\sqrt{x}\sqrt{y-1}}=-1-\frac{1}{x}$$$$\frac{\sqrt{y+1}\sqrt{y-1}+1}{\sqrt{x}\sqrt{y-1}}=\frac{-x-1}{x}$$$${\left(\frac{\sqrt{y+1}\sqrt{y-1}+1}{\sqrt{x}\sqrt{y-1}}\right)}^2={\left(\frac{-x-1}{x}\right)}^2$$$$\frac{y^2-1+1+2\sqrt{y^2-1}}{x(y-1)}=\frac{x^2+2x+1}{x^2}$$$$\frac{y^2+2\sqrt{y^2-1}}{y-1}=\frac{x^2+2x+1}{x}=\frac{xy+2x}{x}=y+2$$$$y^2+2\sqrt{y^2-1}=(y+2)(y-1)=y^2+y-2$$$$2\sqrt{y^2-1}=y-2$$$$4y^2-4=y^2-4y+4$$$$3y^2+4y-8=0$$$$y=\frac{-4\pm \sqrt{16+96}}{6}=\frac{-4\pm \sqrt{112}}{6}$$$$y=\frac{-4\pm 4\sqrt{7}}{6}=\frac{-2\pm 2\sqrt{7}}{3}$$$$\because \frac{-2+2\sqrt{7}}{3}isextraneousroot$$$$\therefore y=\frac{-2-2\sqrt{7}}{3}$$$$\frac{x}{x^2+1}=\frac{1}{y}=\frac{3}{-2-2\sqrt{7}}\cdot \frac{-2+2\sqrt{7}}{-2+2\sqrt{7}}$$$$=\frac{3(-2+2\sqrt{7})}{4-28}=\frac{6(-1+\sqrt{7})}{-24}$$$$=\frac{1-\sqrt{7}}{4}$$

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