For what value of $\;$ $a$ $\;$ is the difference between the roots of the equation $\left(a - 2\right)x^2 - \left(a - 4\right)x - 2 = 0$ equal to $3$?
Given quadratic equation: $\;\;\;$ $\left(a - 2\right)x^2 - \left(a - 4\right)x - 2 = 0$
Relation between the roots $\alpha$ and $\beta$ $\left(\alpha > \beta\right)$ of the given quadratic equation is
$\alpha - \beta = 3$ $\;\;\; \cdots \; (1)$
Sum of roots of the given quadratic equation: $\;$ $\alpha + \beta = \dfrac{a - 4}{a - 2}$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ simultaneously gives
$2 \alpha = 3 + \dfrac{a - 4}{a - 2}$
i.e. $\;$ $2 \alpha = \dfrac{3a - 6 + a - 4}{a - 2}$
i.e. $\;$ $2 \alpha = \dfrac{4a - 10}{a - 2}$
i.e. $\;$ $\alpha = \dfrac{2a - 5}{a - 2}$
Substituting the value of $\alpha$ in equation $(2)$ gives
$\beta = \dfrac{a - 4}{a - 2} - \alpha = \dfrac{a - 4}{a - 2} - \dfrac{2a - 5}{a - 2}$
i.e. $\;$ $\beta = \dfrac{1 - a}{a - 2}$
Product of roots of the given quadratic equation: $\;$ $\alpha \cdot \beta = \dfrac{-2}{a - 2}$ $\;\;\; \cdots \; (3)$
Substituting the values of $\alpha$ and $\beta$ in equation $(3)$ gives
$\left(\dfrac{2a - 5}{a - 2}\right) \left(\dfrac{1 - a}{a - 2}\right) = \dfrac{-2}{a - 2}$
i.e. $\;$ $\left(2a - 5\right) \left(1 - a\right) = -2 \left(a - 2\right)$
i.e. $2a - 2a^2 - 5 + 5a = -2a + 4$
i.e. $\;$ $2a^2 - 9a + 9 = 0$
i.e. $\;$ $\left(a - 3\right) \left(2a - 3\right) = 0$
i.e. $\;$ $a = 3$ $\;$ or $\;$ $a = \dfrac{3}{2}$