Solve the following equation by factorisation:

$x + \dfrac{1}{x} = 2\dfrac{1}{20}$

Given,

$x + \dfrac{1}{x} = 2\dfrac{1}{20} \\[1em] \Rightarrow x \times x + \dfrac{1}{x} \times x = \dfrac{41}{20} \times x \\[1em] \Rightarrow x^2 + 1 = \dfrac{41}{20}x \\[1em] \Rightarrow 20(x^2 + 1) = 41x \\[1em] \Rightarrow 20x^2 + 20 = 41x \\[1em] \Rightarrow 20x^2 - 41x + 20 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 20x^2 - 25x - 16x + 20 = 0 \\[1em] \Rightarrow 5x(4x - 5) - 4(4x - 5) = 0 \\[1em] \Rightarrow (5x - 4)(4x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 5x - 4 = 0 \text{ or } 4x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 5x = 4 \text{ or } 4x = 5 \\[1em] x = \dfrac{4}{5} \text{ or } x = \dfrac{5}{4} \\[1em]$

Hence, the roots of given equation are $\dfrac{4}{5}$ , $\dfrac{5}{4}$.