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Question Number 121073 by Algoritm last updated on 05/Nov/20

Answered by MJS_new last updated on 05/Nov/20

$$\mathrm{both}\mathrm{sides}\mathrm{must}\mathrm{be}\mathrm{greater}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}$$$$\mathrm{zero}\Rightarrow x<0\wedge x\ne -1\vee x>1$$$$\mathrm{squaring}\mathrm{and}\mathrm{transforming}\mathrm{leads}\mathrm{to}$$$$x^5+\frac{x^4}{6}+\frac{x^3}{2}-\frac{x^2}{3}+\frac{1}{6}=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}\frac{1}{6}\mathrm{we}\mathrm{find}$$$$x=-\frac{1}{2}$$$$\mathrm{no}\mathrm{other}\mathrm{solution};\mathrm{the}\mathrm{complex}\mathrm{solutions}\mathrm{of}$$$$\mathrm{the}\mathrm{polynome}\mathrm{do}\mathrm{not}\mathrm{solve}\mathrm{the}\mathrm{given}\mathrm{equation}$$

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