Question

como resolver
(X + 1/X).(X +1/X) = 3

Answer 1

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$\large\texttt{Resolver a equa\c{c}\~ao:}$

$\large\begin{array}{l} \mathtt{\left(x+\dfrac{1}{x}\right)\cdot \left(x+\dfrac{1}{x}\right)=3}\\\\ \mathtt{\left(x+\dfrac{1}{x}\right)^{\!2}=3} \end{array}$


$\large\texttt{Tirando a raiz quadrada dos dois lados, ficamos}\\\texttt{com}$

$\large\begin{array}{l} \mathtt{x+\dfrac{1}{x}=\pm\,\sqrt{3}} \end{array}$


$\large\texttt{Reduzindo o lado esquerdo ao mesmo denominador:}$

$\large\begin{array}{l} \mathtt{\dfrac{x^2}{x}+\dfrac{1}{x}=\pm\,\sqrt{3}}\\\\ \mathtt{\dfrac{x^2+1}{x}=\pm\,\sqrt{3}}\\\\ \mathtt{x^2+1=\pm\,\sqrt{3}\,x}\\\\ \mathtt{x^2\mp\sqrt{3}\,x+1=0}\\\\ \mathtt{x^2-\sqrt{3}\,x+1=0\quad ou\quad x^2+\sqrt{3}\,x+1=0} \end{array}$


$\large\begin{array}{l}\texttt{Temos duas equa\c{c}\~oes quadr\'aticas, mas ambas t\^em}\\\texttt{o mesmo discriminante }\mathtt{\Delta:}\\\\ \left\{\! \begin{array}{l} \mathtt{a=1}\\\mathtt{b=\mp\,\sqrt{3}}\\\mathtt{c=1} \end{array} \right. \end{array}$


$\large\begin{array}{l} \mathtt{\Delta=b^2-4ac}\\\\ \mathtt{\Delta=\big(\!\mp\sqrt{3}\big)^2-4\cdot 1\cdot 1}\\\\ \mathtt{\Delta=3-4}\\\\ \mathtt{\Delta=-1<0} \end{array}$


$\large\begin{array}{l} \texttt{Como o discriminante }\mathtt{\Delta}\texttt{ \'e negativo para ambas as}\\\texttt{equa\c{c}\~oes, nenhuma delas possui solu\c{c}\~ao real.}\\\\\\ \texttt{Sendo assim, o conjunto solu\c{c}\~ao \'e vazio:}\\\\ \mathtt{S=\varnothing.} \end{array}$


$\large\texttt{Bons estudos! :-)}$


Tags:  equação racional quadrática discriminante báscara solução resolver álgebra