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Question Number 57164 by ANTARES VY last updated on 31/Mar/19

$$2\mathbf{x}+1+{\mathbf{x}}^2-{\mathbf{x}}^3+{\mathbf{x}}^4-{\mathbf{x}}^5-.....=\frac{13}{6}$$ $$\mathbf{solved}\mathbf{equation}.$$ $$\mid \mathbf{x}\mid <1$$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Mar/19

$$2x+1+\frac{x^2}{1+x}=\frac{13}{6}$$ $$\frac{2x+2x^2+1+x+x^2}{1+x}=\frac{13}{6}$$ $$18x^2+18x+6=13+13x$$ $$18x^2+5x-7=0$$ $$x=\frac{-5\pm \sqrt{25+4\times 18\times 7}}{2\times 18}$$ $$=\frac{-5\pm \sqrt{529}}{36}$$ $$=\frac{-5\pm 23}{36}\to (\frac{18}{36},\frac{-28}{36})$$ $$x=(\frac{1}{2},\frac{-7}{9})$$

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