$$k^2+2k-8 = \frac{1}{3k^2}+\frac{1}{k^2}$$
Answer
$$ \begin{matrix}k_1 = 2.31191 & k_2 = -5.28203 & k_3 = -0.01494+0.35016i \\[1 em] k_4 = -0.01494-0.35016i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} k^2+2k-8 &= \frac{1}{3k^2}+\frac{1}{k^2}&& \text{multiply ALL terms by } \color{blue}{ 3k^2 }. \\[1 em]3k^2k^2+3k^2\cdot2k-3k^2\cdot8 &= 3k^2\cdot\frac{1}{3k^2}+3k^2\cdot\frac{1}{k^2}&& \text{cancel out the denominators} \\[1 em]3k^4+6k^3-24k^2 &= k^4+3&& \text{move all terms to the left hand side } \\[1 em]3k^4+6k^3-24k^2-k^4-3 &= 0&& \text{simplify left side} \\[1 em]2k^4+6k^3-24k^2-3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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