If $x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}, \text{ find } x + \dfrac{1}{5x}$.

$x^2 + \dfrac{1}{25x^2} = (x)^2 + \dfrac{1}{(5x)^2} ….[i] \\[1em] \Big(x + \dfrac{1}{5x}\Big)^2 = (x)^2 + 2 \times x \times \dfrac{1}{5x} + \dfrac{1}{(5x)^2} \\[1em] = (x)^2 + \dfrac{1}{(5x)^2} + \dfrac{2}{5} \\[1em] \therefore x^2 + \dfrac{1}{25x^2} = \Big(x + \dfrac{1}{5x}\Big)^2 - \dfrac{2}{5}$

Putting this value of $x^2 + \dfrac{1}{25x^2}$ in eqn (i), we get:

$x^2 + \dfrac{1}{25x^2} = \Big(x + \dfrac{1}{5x}\Big)^2 - \dfrac{2}{5}$

Let $x + \dfrac{1}{5x}$ be a.

Substituting values we get,

$\Rightarrow 8\dfrac{3}{5} = a^2 - \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{43}{5} = \dfrac{5a^2 - 2}{5} \\[1em] \Rightarrow 5a^2 - 2 = 43 \\[1em] \Rightarrow 5a^2 = 45 \\[1em] \Rightarrow a^2 = 9 \\[1em] \Rightarrow a = \pm 3.$

Hence, $x + \dfrac{1}{5x} = \pm 3$.