Question Number 194953 by dimentri last updated on 20/Jul/23 $$\:{Let}\:{P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:{and} \\ $$$$\:\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}\:{be}\:{two}\: \\ $$$$\:\:{polynomial}\:{with}\:{real}\:{coefficients} \\ $$$$\:\:{such}\:{that}\:{P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$$\:{for}\:{all}\:{real}\:{x}\:. \\ $$$$\:\:{Find}\:{all}\:{the}\:{real}\:{roots}\:{of}\: \\ $$$$\:\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\: \\…
Question Number 194952 by pascal889 last updated on 20/Jul/23 $$\boldsymbol{\mathrm{x}}\:+\:\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\:+\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{X}} \\ $$ Answered by Frix last updated on 20/Jul/23 $$\mathrm{Obviously}\:{x}=\mathrm{1}\vee{x}=\mathrm{4} \\…
Question Number 194900 by cortano12 last updated on 19/Jul/23 $$\:\:\:\:\:\:{Given}\:\:\:{d}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\:\sqrt{\mathrm{5}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:\: \\ $$$$\:\:\:\:\:\:{then}\:\:{d}^{\mathrm{3}} −\mathrm{4}{d}^{\mathrm{2}} \:+\mathrm{8}{d}\:−\mathrm{2}\:=?\: \\ $$$$\:\:\:\:\: \\ $$ Answered by Frix last updated on 19/Jul/23…
Question Number 194899 by kapoorshah last updated on 19/Jul/23 Answered by Frix last updated on 19/Jul/23 $${y}={px}\wedge{z}={qx} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{1}−{pq}}=\frac{\mathrm{8}}{{p}^{\mathrm{2}} −{q}}=\frac{\mathrm{10}}{{q}^{\mathrm{2}} −{p}} \\…
Question Number 194914 by Mingma last updated on 19/Jul/23 Answered by a.lgnaoui last updated on 20/Jul/23 $$\mathrm{distance}\:\mathrm{betwen}\:\mathrm{Dand}\:\mathrm{E}\:=\mathrm{distance}\:\mathrm{betwen}\: \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{D};\:\:\:\mathrm{so}\:\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{point} \\ $$$$\:\mathrm{4}^{\mathrm{9}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{A}} \\ $$$$ \\…
Question Number 194877 by SajaRashki last updated on 18/Jul/23 $${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\ $$ Commented by SajaRashki last updated on 24/Jul/23 $${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\…
Question Number 194888 by Abdullahrussell last updated on 18/Jul/23 Answered by aleks041103 last updated on 18/Jul/23 $${The}\:{combined}\:{ages}\:{of}\:{Mary}\:{and}\:{Ann}\:{is}\:\mathrm{44}\:{years}. \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}} \right)+{A}\left({t}_{\mathrm{1}} \right)=\mathrm{44}\:\right) \\ $$$${Mary}\:{is}\:{twice}\:{as}\:{old}\:{as}\:{Ann}\:{was} \\ $$$$\left(\:{M}\left({t}_{\mathrm{1}}…
Question Number 194884 by cortano12 last updated on 18/Jul/23 $$\:\:\:\:\:\:{x}!\:=\:\mathrm{6}!.\:\mathrm{7}!\: \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:=?\: \\ $$ Answered by AST last updated on 18/Jul/23 $$=\mathrm{7}!×\left(\mathrm{2}×\mathrm{3}\right)×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}= \\ $$$$\mathrm{7}!×\left(\mathrm{4}×\mathrm{2}\right)×\left(\mathrm{3}×\mathrm{3}\right)×\left(\mathrm{5}×\mathrm{2}\right)=\mathrm{10}!={x}!\Rightarrow{x}^{\mathrm{2}}…
Question Number 194846 by dimentri last updated on 17/Jul/23 $$\:\:\:\:\:\:\underbrace{\:} \\ $$ Answered by som(math1967) last updated on 17/Jul/23 $$\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}+\mathrm{6}\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{3}.\mathrm{3}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}}…
Question Number 194853 by horsebrand11 last updated on 17/Jul/23 Answered by cortano12 last updated on 17/Jul/23 $$\:\:\:\underbrace{\Subset} \\ $$ Commented by horsebrand11 last updated on…