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 90212 by john santu last updated on 22/Apr/20

if x^2 = 5x+1 find E = (((x^3 −140) (x^(11) )^(1/(3 )) )/((x^4 +1))^(1/(3 )) )

$${if}\:{x}^{\mathrm{2}} \:=\:\mathrm{5}{x}+\mathrm{1}\: \\ $$$${find}\:{E}\:=\:\frac{\left({x}^{\mathrm{3}} −\mathrm{140}\right)\:\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{11}} }}{\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{4}} +\mathrm{1}}} \\ $$

Commented by MJS last updated on 22/Apr/20

(1/3)

$$\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by jagoll last updated on 22/Apr/20

how sir?

$$\mathrm{how}\:\mathrm{sir}? \\ $$

Commented by MJS last updated on 22/Apr/20

E^3 =(((x^3 −140)^3 x^(11) )/((x^4 +1)))=(((x^2 x−140)^3 (x^2 )^5 x)/(((x^2 )^2 +1))) x^2 =5x+1 E^3 =((x(5x+1)^5 (5x^2 +x−140)^3 )/(25x^2 +10x+2)) x^2 −5x−1=0 ⇒ x=((5±(√(29)))/2) x=(5/2)+(v/2) E^3 =(((v+5)(5v+27)^5 (5v^2 +52v−425)^3 )/(1024(25v^2 +270v+733))) v^2 =29 E^3 =(((v+5)(5v+27)^5 (52v−280))/(1024(270v+1458)))= =(((v+5)(5v+27)^5 4(13v−70))/(1024×54(5v+27)))= =(1/(864))(v+5)(5v+27)^4 (13v−70)^3 = =(1/(864))((v+5)(13v−70))((5v+27)^2 )^2 (13v−70)^2 = =(1/(864))(13v^2 −5v−350)(25v^2 +270v+729)^2 (169v^2 −1820v+4900)= v^2 =29 =(1/(864))(27−5v)(270v+1454)^2 (9801−1820v)= =(1/(216))(5v−27)(135v+727)^2 (1820v−9801)= =(1/(216))(5v−27)(18225v^2 +196290v+528529)(1820v−9801)= v^2 =29 (1/(216))(5v−27)(196290v+1057054)(1820v−9801)= =(1/(108))(5v−27)(1820v−9801)(98145v+528527)= =(1/(108))(9100v^2 −98145v+264627)(98145v+528527)= v^2 =29 (1/(108))(−98145v+528527)(98145v+528527)= =(1/(108))(−9632441025v^2 +279340789729)= v^2 =29 =(1/(108))(4)=(1/(27)) ⇒ E=(1/3)

$${E}^{\mathrm{3}} =\frac{\left({x}^{\mathrm{3}} −\mathrm{140}\right)^{\mathrm{3}} {x}^{\mathrm{11}} }{\left({x}^{\mathrm{4}} +\mathrm{1}\right)}=\frac{\left({x}^{\mathrm{2}} {x}−\mathrm{140}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} \right)^{\mathrm{5}} {x}}{\left(\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${x}^{\mathrm{2}} =\mathrm{5}{x}+\mathrm{1} \\ $$$${E}^{\mathrm{3}} =\frac{{x}\left(\mathrm{5}{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{5}{x}^{\mathrm{2}} +{x}−\mathrm{140}\right)^{\mathrm{3}} }{\mathrm{25}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{29}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}}+\frac{{v}}{\mathrm{2}} \\ $$$${E}^{\mathrm{3}} =\frac{\left({v}+\mathrm{5}\right)\left(\mathrm{5}{v}+\mathrm{27}\right)^{\mathrm{5}} \left(\mathrm{5}{v}^{\mathrm{2}} +\mathrm{52}{v}−\mathrm{425}\right)^{\mathrm{3}} }{\mathrm{1024}\left(\mathrm{25}{v}^{\mathrm{2}} +\mathrm{270}{v}+\mathrm{733}\right)} \\ $$$${v}^{\mathrm{2}} =\mathrm{29} \\ $$$${E}^{\mathrm{3}} =\frac{\left({v}+\mathrm{5}\right)\left(\mathrm{5}{v}+\mathrm{27}\right)^{\mathrm{5}} \left(\mathrm{52}{v}−\mathrm{280}\right)}{\mathrm{1024}\left(\mathrm{270}{v}+\mathrm{1458}\right)}= \\ $$$$=\frac{\left({v}+\mathrm{5}\right)\left(\mathrm{5}{v}+\mathrm{27}\right)^{\mathrm{5}} \mathrm{4}\left(\mathrm{13}{v}−\mathrm{70}\right)}{\mathrm{1024}×\mathrm{54}\left(\mathrm{5}{v}+\mathrm{27}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{864}}\left({v}+\mathrm{5}\right)\left(\mathrm{5}{v}+\mathrm{27}\right)^{\mathrm{4}} \left(\mathrm{13}{v}−\mathrm{70}\right)^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{864}}\left(\left({v}+\mathrm{5}\right)\left(\mathrm{13}{v}−\mathrm{70}\right)\right)\left(\left(\mathrm{5}{v}+\mathrm{27}\right)^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{13}{v}−\mathrm{70}\right)^{\mathrm{2}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{864}}\left(\mathrm{13}{v}^{\mathrm{2}} −\mathrm{5}{v}−\mathrm{350}\right)\left(\mathrm{25}{v}^{\mathrm{2}} +\mathrm{270}{v}+\mathrm{729}\right)^{\mathrm{2}} \left(\mathrm{169}{v}^{\mathrm{2}} −\mathrm{1820}{v}+\mathrm{4900}\right)= \\ $$$${v}^{\mathrm{2}} =\mathrm{29} \\ $$$$=\frac{\mathrm{1}}{\mathrm{864}}\left(\mathrm{27}−\mathrm{5}{v}\right)\left(\mathrm{270}{v}+\mathrm{1454}\right)^{\mathrm{2}} \left(\mathrm{9801}−\mathrm{1820}{v}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{216}}\left(\mathrm{5}{v}−\mathrm{27}\right)\left(\mathrm{135}{v}+\mathrm{727}\right)^{\mathrm{2}} \left(\mathrm{1820}{v}−\mathrm{9801}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{216}}\left(\mathrm{5}{v}−\mathrm{27}\right)\left(\mathrm{18225}{v}^{\mathrm{2}} +\mathrm{196290}{v}+\mathrm{528529}\right)\left(\mathrm{1820}{v}−\mathrm{9801}\right)= \\ $$$${v}^{\mathrm{2}} =\mathrm{29} \\ $$$$\frac{\mathrm{1}}{\mathrm{216}}\left(\mathrm{5}{v}−\mathrm{27}\right)\left(\mathrm{196290}{v}+\mathrm{1057054}\right)\left(\mathrm{1820}{v}−\mathrm{9801}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{108}}\left(\mathrm{5}{v}−\mathrm{27}\right)\left(\mathrm{1820}{v}−\mathrm{9801}\right)\left(\mathrm{98145}{v}+\mathrm{528527}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{108}}\left(\mathrm{9100}{v}^{\mathrm{2}} −\mathrm{98145}{v}+\mathrm{264627}\right)\left(\mathrm{98145}{v}+\mathrm{528527}\right)= \\ $$$${v}^{\mathrm{2}} =\mathrm{29} \\ $$$$\frac{\mathrm{1}}{\mathrm{108}}\left(−\mathrm{98145}{v}+\mathrm{528527}\right)\left(\mathrm{98145}{v}+\mathrm{528527}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{108}}\left(−\mathrm{9632441025}{v}^{\mathrm{2}} +\mathrm{279340789729}\right)= \\ $$$${v}^{\mathrm{2}} =\mathrm{29} \\ $$$$=\frac{\mathrm{1}}{\mathrm{108}}\left(\mathrm{4}\right)=\frac{\mathrm{1}}{\mathrm{27}} \\ $$$$\Rightarrow\:{E}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by jagoll last updated on 22/Apr/20

amazing sir. thank you

$$\mathrm{amazing}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com