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Question-143731




Question Number 143731 by haladu last updated on 17/Jun/21
Answered by MJS_new last updated on 17/Jun/21
if we find x>0 with ((2x+1)/(x−1))=x^4  we have a real  solution  ⇒  x^5 −x^4 −2x−1=0  (x^2 −x−1)(x^3 +x+1)=0  ⇒ x=((1+(√5))/2)
$$\mathrm{if}\:\mathrm{we}\:\mathrm{find}\:{x}>\mathrm{0}\:\mathrm{with}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{1}}={x}^{\mathrm{4}} \:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{real} \\ $$$$\mathrm{solution} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{5}} −{x}^{\mathrm{4}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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