Solve the following equations by using formula:

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

Given,

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

$\Rightarrow \dfrac{x^2 - 1}{x} = 3 \text{ (By taking L.C.M) } \\[0.5em] \Rightarrow x^2 - 1 = 3x \\[0.5em] \Rightarrow x^2 - 3x - 1 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -3 , c = -1

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-3) ± \sqrt{(-3)^2 - 4 \times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{9 - (-4)}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{13}}{2} \text{ or } \dfrac{3 - \sqrt{13}}{2} \\[1em]$

Hence roots of the given equations are $\dfrac{3 + \sqrt{13}}{2} , \dfrac{3 - \sqrt{13}}{2}$.