Question

calcule a soma dos inversos das raizes da equação x^3-5x^2+4=0

Answer 1

$\mathrm{a_3x^3+a_2x^2+a_1x+a_0=0\ \to\ x^3-5x^2+4=0}\\ \mathrm{a_3=1\ \ \| \ \ a_2=-5\ \ \| \ \ a_1=0\ \ \| \ \ a_0=4}\\\\ \textbf{Atrav\'es do teorema da raiz racional, teremos que:}\\\\ \mathrm{*\ Poss\'iveis\ ra\'izes\ \to\ \pm\dfrac{divisores\ de\ a_0}{divisores\ de\ a_n}\ |\ a_0\ e\ a_n\in\mathbb{Z}}}}\\\\ \mathrm{*\ Divisores\ de\ a_0\ \to\ 1,2,4\ \ \| \ \ *\ Divisores\ de\ a_3\ \to\ 1}\\\\ \mathrm{*\ Poss\'iveis\ ra\'izes\ \to\ \pm1,\pm2,\pm4}}}\\\\ \mathrm{*\ Para\ x=1\ \to\ 1^3-5.1^2+4=0\ \to\ Ok\ \to\ x_1=1}$

$\textbf{Fatorando o polin\^omio, teremos que:}\\\\ \mathrm{(x-x_1)\dfrac{P(x)}{(x-x_1)}=0\ \to\ (x-1)\dfrac{x^3-5x^2+4}{(x-1)}=0}\\\\ \mathrm{(x-1)(x^2-4x-4)=0\ \to\ x^2-4x-4=0}\\\\ \mathrm{x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4.1(-4)}}{2.1}=\dfrac{4\pm\sqrt{32}}{2}=}\\\\ \mathrm{=\dfrac{4\pm\sqrt{2^5}}{2}=\dfrac{4\pm4\sqrt{2}}{2}=2\pm2\sqrt{2}}\\\\ \mathrm{x'=2+2\sqrt{2}=2(1+\sqrt{2})\ \ \| \ \ x''=2-2\sqrt{2}=2(1-\sqrt{2})}$

$\textbf{Portanto, as solu\c{c}\~oes para}\ \mathbf{x^3-5x^2+4=0\ s\~ao:}\\\\ \mathrm{S=\{x_1,x_2,x_3\}=\{1,2(1+\sqrt{2}),2(1-\sqrt{2})\}}\\\\ \textbf{Calculando a soma dos inversos das ra\'izes:}\\\\ \mathrm{\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}=\dfrac{1}{1}+\dfrac{1}{2(1+\sqrt{2})}+\dfrac{1}{2(1-\sqrt{2})}=}\\\\\\ \mathrm{=1+\dfrac{1}{2}\Bigg(\dfrac{1}{(1+\sqrt{2})}+\dfrac{1}{(1-\sqrt{2})}\Bigg)=}$

$\mathrm{=1+\dfrac{1}{2}\Bigg(\dfrac{1-\sqrt{2}+1+\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}\Bigg)=1+\dfrac{1}{2}\Bigg(\dfrac{2}{1-2}\Bigg)=}\\\\\\ \mathrm{=1+\dfrac{1}{2}\Bigg(\dfrac{2}{-1}\Bigg)=1+(-1)=1-1=\mathbf{0}}$