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Question Number 185210 by mathlove last updated on 18/Jan/23
x=((x^2 +1)/(α+1)) x=?
$${x}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\alpha+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$
Commented by Frix last updated on 18/Jan/23
Simply transform it to x^2 +px+q=0 and solve it. What′s the problem? Laziness?
$$\mathrm{Simply}\:\mathrm{transform}\:\mathrm{it}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{solve}\:\mathrm{it}. \\ $$$$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{problem}?\:\mathrm{Laziness}? \\ $$
Commented by aba last updated on 18/Jan/23
if x^2 +px+q=0 x=((−p±(√(p^2 −4q)))/2) the two root are : x_1 =((−p+(√(p^2 −4q)))/2) and x_2 =((−p−(√(p^2 −4q)))/2) so x_1 +x_2 =(√(p^2 −4q))=(√Δ)
$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} +\mathrm{px}+\mathrm{q}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{−\mathrm{p}\pm\sqrt{\mathrm{p}^{\mathrm{2}} −\mathrm{4q}}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{root}\:\mathrm{are}\:: \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{−\mathrm{p}+\sqrt{\mathrm{p}^{\mathrm{2}} −\mathrm{4q}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{−\mathrm{p}−\sqrt{\mathrm{p}^{\mathrm{2}} −\mathrm{4q}}}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\sqrt{\mathrm{p}^{\mathrm{2}} −\mathrm{4q}}=\sqrt{\Delta} \\ $$
Answered by aba last updated on 18/Jan/23
x=(1/2)(a+1±(√(a^2 +2a−3)))
$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{1}\pm\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{2a}−\mathrm{3}}\right)\: \\ $$$$ \\ $$
Commented by Frix last updated on 18/Jan/23
Yes. Now please solve this for me, I am so very lazy today too: ((x^2 +2γ)/(γ−δx))=1
$$\mathrm{Yes}.\:\mathrm{Now}\:\mathrm{please}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:\mathrm{me},\:\mathrm{I}\:\mathrm{am} \\ $$$$\mathrm{so}\:\mathrm{very}\:\mathrm{lazy}\:\mathrm{today}\:\mathrm{too}: \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{2}\gamma}{\gamma−\delta{x}}=\mathrm{1} \\ $$
Commented by aba last updated on 18/Jan/23
x=±(1/2)(√(δ^2 −4γ))−(δ/2)
$$\mathrm{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\delta^{\mathrm{2}} −\mathrm{4}\gamma}−\frac{\delta}{\mathrm{2}} \\ $$
Commented by Frix last updated on 18/Jan/23
Thank you. But what about this: ((y^2 +2δ)/(δ−εy))=1
$$\mathrm{Thank}\:\mathrm{you}.\:\mathrm{But}\:\mathrm{what}\:\mathrm{about}\:\mathrm{this}: \\ $$$$\frac{{y}^{\mathrm{2}} +\mathrm{2}\delta}{\delta−\epsilon{y}}=\mathrm{1} \\ $$
Commented by aba last updated on 18/Jan/23
??
$$?? \\ $$
Answered by manxsol last updated on 18/Jan/23
Seeing Beyond the Obvious x+(1/x)=α+1 analisis x+(1/x)≥2 ∀xeR MA≥MG⇒x+(1/x)≥2(√(x.(1/x))) x+(1/x)≥2 α+1≥2 α≥1 x^2 +2+(1/x^2 )=(α+1)^2 x^2 −2+(1/x^2 )=(α+1)^2 −4 (x−(1/x))^2 =(α+3)(α−1) x−(1/x)=±(√((α+3)(α−1))) x+(1/x)=α+1 2x=α+1±(√((α+3)(α−1))) x_1 =((α+1+(√((α+3)(α−1))))/2) x_2 =((α+1−(√((α+3)(α−1))))/2) check condiciones iniciales α≥1⇒α−1≥0✓ ok roots ik
$$ \\ $$$${Seeing}\:{Beyond}\:{the}\:{Obvious} \\ $$$$ \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\alpha+\mathrm{1} \\ $$$${analisis} \\ $$$${x}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2}\:\:\:\forall{xeR} \\ $$$${MA}\geqslant{MG}\Rightarrow{x}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2}\sqrt{{x}.\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\frac{\mathrm{1}}{{x}}\geqslant\mathrm{2} \\ $$$$\alpha+\mathrm{1}\geqslant\mathrm{2} \\ $$$$\alpha\geqslant\mathrm{1} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\alpha+\mathrm{3}\right)\left(\alpha−\mathrm{1}\right) \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\pm\sqrt{\left(\alpha+\mathrm{3}\right)\left(\alpha−\mathrm{1}\right)} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\alpha+\mathrm{1} \\ $$$$\mathrm{2}{x}=\alpha+\mathrm{1}\pm\sqrt{\left(\alpha+\mathrm{3}\right)\left(\alpha−\mathrm{1}\right)} \\ $$$${x}_{\mathrm{1}} =\frac{\alpha+\mathrm{1}+\sqrt{\left(\alpha+\mathrm{3}\right)\left(\alpha−\mathrm{1}\right)}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{\alpha+\mathrm{1}−\sqrt{\left(\alpha+\mathrm{3}\right)\left(\alpha−\mathrm{1}\right)}}{\mathrm{2}} \\ $$$${check}\:{condiciones}\:{iniciales} \\ $$$$\alpha\geqslant\mathrm{1}\Rightarrow\alpha−\mathrm{1}\geqslant\mathrm{0}\checkmark \\ $$$${ok}\:{roots} \\ $$$$ \\ $$$${ik} \\ $$
Answered by MJS_new last updated on 18/Jan/23
just because I also want to post something let α=((ζ^2 −ζ+1)/ζ) ⇒ x=((ζ(x^2 +1))/(ζ^2 +1)) ⇔ x^2 −((ζ^2 +1)/ζ)x+1=0 ⇒ x=ζ∨x=(1/ζ) which I like more than the simple answers you gave
$$\mathrm{just}\:\mathrm{because}\:\mathrm{I}\:\mathrm{also}\:\mathrm{want}\:\mathrm{to}\:\mathrm{post}\:\mathrm{something} \\ $$$$\mathrm{let}\:\alpha=\frac{\zeta^{\mathrm{2}} −\zeta+\mathrm{1}}{\zeta} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\zeta\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\zeta^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:{x}^{\mathrm{2}} −\frac{\zeta^{\mathrm{2}} +\mathrm{1}}{\zeta}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}=\zeta\vee{x}=\frac{\mathrm{1}}{\zeta} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{like}\:\mathrm{more}\:\mathrm{than}\:\mathrm{the}\:\mathrm{simple}\:\mathrm{answers} \\ $$$$\mathrm{you}\:\mathrm{gave} \\ $$
Commented by manxsol last updated on 19/Jan/23
i liked your reasoning. for me toolnox
$${i}\:{liked}\:{your}\:{reasoning}. \\ $$$$\:{for}\:{me}\:{toolnox} \\ $$

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