Using the properties of proportion, solve for x,

given $\dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8}$

Given,

$\Rightarrow \dfrac{x^4 + 1}{2x^2} = \dfrac{17}{8}$

Applying componendo and dividendo, we get

$\Rightarrow \dfrac{x^4 + 1 + 2x^2}{x^4 + 1 - 2x^2} = \dfrac{17 + 8}{17 - 8} \\[1em] \Rightarrow \dfrac{(x^2 + 1)^2}{(x^2 - 1)^2} = \dfrac{25}{9} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x^2 - 1} = \dfrac{5}{3} \\[1em]$

Applying componendo and dividendo, we get

$\Rightarrow \dfrac{x^2 + 1 + x^2 - 1}{x^2 + 1 - (x^2 - 1)} = \dfrac{5 + 3}{5 - 3} \\[1em] \Rightarrow \dfrac{2x^2}{2} = \dfrac{8}{2} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x = \pm 2.$

Hence, x = ±2.