If m and n are roots of the equation :

$\dfrac{1}{x} - \dfrac{1}{x - 2} = 3$; where x ≠ 0 and x ≠ 2; find m × n.

Solving,

$\Rightarrow \dfrac{1}{x} - \dfrac{1}{x - 2} = 3 \\[1em] \dfrac{x - 2 - x}{x(x - 2)} = 3 \\[1em] \dfrac{-2}{x(x - 2)} = 3 \\[1em] -2 = 3x(x - 2) \\[1em] -2 = 3x^2 - 6x \\[1em] 3x^2 - 6x + 2 = 0 \\[1em]$

Comparing 3x² - 6x + 2 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -6 and c = 2.

We know that,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \\[1em] x = \dfrac{6 \pm \sqrt{36 - 24}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{12}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{4 \times 3}}{6} \\[1em] x = \dfrac{6 \pm 2\sqrt{3}}{6} \\[1em] x = \dfrac{2(3 \pm \sqrt{3})}{6} \\[1em] x = \dfrac{3 + \sqrt{3}}{3}, \dfrac{3 - \sqrt{3}}{3} \\[1em] \therefore m = \dfrac{3 + \sqrt{3}}{3} \text{ and } n = \dfrac{3 - \sqrt{3}}{3}.$

Solving m × n,

$m × n = \Big(\dfrac{3 + \sqrt{3}}{3}\Big)\Big(\dfrac{3 - \sqrt{3}}{3}\Big) \\[1em] = \dfrac{(3)^2 - (\sqrt{3})^2}{9} \\[1em] = \dfrac{9 - 3}{9} \\[1em] = \dfrac{6}{9} \\[1em] = \dfrac{2}{3}.$

Hence, m × n = $\dfrac{2}{3}$.