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 126329 by mnjuly1970 last updated on 19/Dec/20

Answered by Olaf last updated on 19/Dec/20

x^5 −5x^4 −10x^3 +10x^2 +5x−1 = 0  (x^5 −1)−5(x^4 −x)−10(x^3 −x^2 ) = 0  x = 1 or :  (x^4 +x^3 +x^2 +x+1)−5x(x^2 +x+1)−10x^2  = 0  x^4 −4x^3 −14x^2 −4x+1 = 0  x^2 −4x−14−(4/x)+(1/x^2 ) = 0  (x^2 +(1/x^2 ))−4(x+(1/x))−14 = 0  (x+(1/x))^2 −2−4(x+(1/x))−14 = 0  (x+(1/x))^2 −4(x+(1/x))−16 = 0  Let u = x+(1/x)  u^2 −4u−16 = 0  u = ((4±(√(16−4(1)(−16))))/2) = 2(1±(√5))  x+(1/x)= u  x^2 −ux+1 = 0  x = ((u±(√(u^2 −4)))/2)  If u = 2(1−(√5)), u^2 −4 = 4(5−2(√5))  x = 1−(√5)±2(√(5−2(√5)))  If u = 2(1+(√5)), u^2 −4 = 4(5+2(√5))  x = 1+(√5)±2(√(5+2(√5))))

$${x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({x}^{\mathrm{5}} −\mathrm{1}\right)−\mathrm{5}\left({x}^{\mathrm{4}} −{x}\right)−\mathrm{10}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{1}\:\mathrm{or}\:: \\ $$$$\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{5}{x}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{10}{x}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{14}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{14}−\frac{\mathrm{4}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\mathrm{4}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{14}\:=\:\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}−\mathrm{4}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{14}\:=\:\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{16}\:=\:\mathrm{0} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{1}}{{x}} \\ $$$${u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{16}\:=\:\mathrm{0} \\ $$$${u}\:=\:\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{16}\right)}}{\mathrm{2}}\:=\:\mathrm{2}\left(\mathrm{1}\pm\sqrt{\mathrm{5}}\right) \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\:{u} \\ $$$${x}^{\mathrm{2}} −{ux}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{If}\:{u}\:=\:\mathrm{2}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right),\:{u}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{4}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}\right) \\ $$$${x}\:=\:\mathrm{1}−\sqrt{\mathrm{5}}\pm\mathrm{2}\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\mathrm{If}\:{u}\:=\:\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right),\:{u}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{4}\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right) \\ $$$${x}\:=\:\mathrm{1}+\sqrt{\mathrm{5}}\pm\mathrm{2}\sqrt{\left.\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}\right)} \\ $$

Commented by mnjuly1970 last updated on 19/Dec/20

mercey mr olaf...thank you...

$${mercey}\:{mr}\:{olaf}...{thank}\:{you}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com