Solution
$$5x + \frac{1}{3x} = 4$$
$$\frac{3}{5}(5x + \frac{1}{3x}) = \frac{3}{5} \times 4$$
$$(3x + \frac{1}{5x}) = \frac{12}{5}$$
On square both sides,
$$(3x + \frac{1}{5x})^2 = (\frac{12}{5})^2$$
$$9x^2 + \frac{1}{25x^2} + 2.3x.\frac{1}{5x} = \frac{144}{25}$$
$$((a+b)^2 = a^2 + b^2 + 2ab)$$
$$9x^2 + \frac{1}{25x^2} = \frac{144}{25} - \frac{6}{5} = 0$$
$$9x^2 + \frac{1}{25x^2} = \frac{114}{25}$$
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