Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 185210 by mathlove last updated on 18/Jan/23

$$x=\frac{x^2+1}{\alpha +1}x=?$$

Commented by Frix last updated on 18/Jan/23

$$\mathrm{Simply}\mathrm{transform}\mathrm{it}\mathrm{to}$$$$x^2+px+q=0$$$$\mathrm{and}\mathrm{solve}\mathrm{it}.$$$${\mathrm{What}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{problem}?\mathrm{Laziness}?$$

Commented by aba last updated on 18/Jan/23

$$\mathrm{if}{\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0$$$$\mathrm{x}=\frac{-\mathrm{p}\pm \sqrt{{\mathrm{p}}^2-4\mathrm{q}}}{2}$$$$\mathrm{the}\mathrm{two}\mathrm{root}\mathrm{are}:$$$${\mathrm{x}}_1=\frac{-\mathrm{p}+\sqrt{{\mathrm{p}}^2-4\mathrm{q}}}{2}\mathrm{and}{\mathrm{x}}_2=\frac{-\mathrm{p}-\sqrt{{\mathrm{p}}^2-4\mathrm{q}}}{2}$$$$\mathrm{so}{\mathrm{x}}_1+{\mathrm{x}}_2=\sqrt{{\mathrm{p}}^2-4\mathrm{q}}=\sqrt{\mathrm{\Delta }}$$

Answered by aba last updated on 18/Jan/23

$$\mathrm{x}=\frac{1}{2}(\mathrm{a}+1\pm \sqrt{{\mathrm{a}}^2+2\mathrm{a}-3})$$$$$$

Commented by Frix last updated on 18/Jan/23

$$\mathrm{Yes}.\mathrm{Now}\mathrm{please}\mathrm{solve}\mathrm{this}\mathrm{for}\mathrm{me},\mathrm{I}\mathrm{am}$$$$\mathrm{so}\mathrm{very}\mathrm{lazy}\mathrm{today}\mathrm{too}:$$$$\frac{x^2+2\gamma }{\gamma -\delta x}=1$$

Commented by aba last updated on 18/Jan/23

$$\mathrm{x}=\pm \frac{1}{2}\sqrt{{\delta }^2-4\gamma }-\frac{\delta }{2}$$

Commented by Frix last updated on 18/Jan/23

$$\mathrm{Thank}\mathrm{you}.\mathrm{But}\mathrm{what}\mathrm{about}\mathrm{this}:$$$$\frac{y^2+2\delta }{\delta -ϵy}=1$$

Commented by aba last updated on 18/Jan/23

$$??$$

Answered by manxsol last updated on 18/Jan/23

$$$$$$SeeingBeyondtheObvious$$$$$$$$x+\frac{1}{x}=\alpha +1$$$$analisis$$$$x+\frac{1}{x}⩾2\mathrm{\forall }xeR$$$$MA⩾MG\Rightarrow x+\frac{1}{x}⩾2\sqrt{x.\frac{1}{x}}$$$$x+\frac{1}{x}⩾2$$$$\alpha +1⩾2$$$$\alpha ⩾1$$$$$$$$x^2+2+\frac{1}{x^2}={(\alpha +1)}^2$$$$x^2-2+\frac{1}{x^2}={(\alpha +1)}^2-4$$$${(x-\frac{1}{x})}^2=(\alpha +3)(\alpha -1)$$$$x-\frac{1}{x}=\pm \sqrt{(\alpha +3)(\alpha -1)}$$$$x+\frac{1}{x}=\alpha +1$$$$2x=\alpha +1\pm \sqrt{(\alpha +3)(\alpha -1)}$$$$x_1=\frac{\alpha +1+\sqrt{(\alpha +3)(\alpha -1)}}{2}$$$$x_2=\frac{\alpha +1-\sqrt{(\alpha +3)(\alpha -1)}}{2}$$$$checkcondicionesiniciales$$$$\alpha ⩾1\Rightarrow \alpha -1⩾0✓$$$$okroots$$$$$$$$ik$$

Answered by MJS_new last updated on 18/Jan/23

$$\mathrm{just}\mathrm{because}\mathrm{I}\mathrm{also}\mathrm{want}\mathrm{to}\mathrm{post}\mathrm{something}$$$$\mathrm{let}\alpha =\frac{{\zeta }^2-\zeta +1}{\zeta }$$$$\Rightarrow$$$$x=\frac{\zeta (x^2+1)}{{\zeta }^2+1}\iff x^2-\frac{{\zeta }^2+1}{\zeta }x+1=0$$$$\Rightarrow$$$$x=\zeta \vee x=\frac{1}{\zeta }$$$$\mathrm{which}\mathrm{I}\mathrm{like}\mathrm{more}\mathrm{than}\mathrm{the}\mathrm{simple}\mathrm{answers}$$$$\mathrm{you}\mathrm{gave}$$

Commented by manxsol last updated on 19/Jan/23

$$ilikedyourreasoning.$$$$formetoolnox$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com