$$\frac{18}{x^4}+\frac{1}{x^2}-4 = 0$$
Answer
$$ \begin{matrix}x_1 = -1.21887 & x_2 = 1.21887 & x_3 = 0.84487+0.86213i \\[1 em] x_4 = 0.84487-0.86213i & x_5 = -0.84487+0.86213i & x_6 = -0.84487-0.86213i \\[1 em] x_7 = 1.19445i & x_8 = -1.19445i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{18}{x^4}+\frac{1}{x^2}-4 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^4x^2 }. \\[1 em]x^4x^2\cdot\frac{18}{x^4}+x^4x^2\cdot\frac{1}{x^2}-x^4x^2\cdot4 &= x^4x^2\cdot0&& \text{cancel out the denominators} \\[1 em]\frac{18}{x^2}+1-4x^6 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2 }. \\[1 em]x^2\cdot\frac{18}{x^2}+x^2\cdot1-x^2\cdot4x^6 &= x^2\cdot0&& \text{cancel out the denominators} \\[1 em]18+x^2-4x^8 &= 0&& \text{simplify left side} \\[1 em]-4x^8+x^2+18 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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