Question Number 195002 by mathlove last updated on 22/Jul/23 $${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$ Answered by dimentri last updated on 22/Jul/23 $$\:\:\:\left({x}\right)^{\sqrt{{x}}} \:=\:\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} \\ $$$$\:\:\left(\mathrm{1}\right)\:\sqrt{{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\Rightarrow{x}−\mathrm{2}\sqrt{{x}}\:=\mathrm{0}…
Question Number 195027 by York12 last updated on 22/Jul/23 $${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}}…
Question Number 195017 by mathlove last updated on 22/Jul/23 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 22/Jul/23 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$$${Let}\:\:\:\:\:{x}+\mathrm{1}={y} \\…
Question Number 194991 by York12 last updated on 21/Jul/23 $$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}}…
Question Number 194942 by horsebrand11 last updated on 20/Jul/23 $$\:\:\:\:\: \\ $$ Commented by MJS_new last updated on 20/Jul/23 $$−\mathrm{2} \\ $$ Commented by horsebrand11…
Question Number 194953 by dimentri last updated on 20/Jul/23 $$\:{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}\: \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{D};\:\:\:\mathrm{so}\:\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{point} \\ $$$$\:\mathrm{4}^{\mathrm{9}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\boldsymbol{\mathrm{A}} \\ $$$$ \\…