Question
$$(x+\frac{1}{x})(x-\frac{2}{x}) = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} (x+\frac{1}{x})(x-\frac{2}{x}) &= 0&& \text{simplify left side} \\[1 em]\frac{x^2+1}{x}\frac{x^2-2}{x} &= 0&& \\[1 em]\frac{x^4-x^2-2}{x^2} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2 }. \\[1 em]x^2\frac{x^4-x^2-2}{x^2} &= x^2\cdot0&& \text{cancel out the denominators} \\[1 em]x^8-x^6-2x^4 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{8}-x^{6}-2x^{4} = 0 } $, first we need to factor our $ x^4 $.
$$ x^{8}-x^{6}-2x^{4} = x^4 \left( x^{4}-x^{2}-2 \right) $$$ x = 0 $ is a root of multiplicity $ 4 $.
The remaining roots can be found by solving equation $ x^{4}-x^{2}-2 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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