If x - 3 = $\dfrac{1}{x}$, find the value of $x^2 + \dfrac{1}{x^2}.$

Given,

$\phantom{\therefore} x - 3 = \dfrac{1}{x} \\[1em] \therefore x - \dfrac{1}{x} = 3 \\[1em]$

We know that,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)^2 + 2.$

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = (3)^2 + 2 \\[1em] = 9 + 2 \\[1em] = 11$

Hence, $x^2 + \dfrac{1}{x^2}$ = 11