Let $\lim\limits_{x \to 2} \dfrac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in \mathbb{R}$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

Given limit is $\lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} \cdot \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5$

Since $\lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} = 1$, we have:

$\lim\limits_{x \to 2} \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5$

For the limit to exist, the numerator must be zero at $x = 2$:

$r(2)^2 + (p-2)(2) - 2p = 0$

$4r + 2p - 4 - 2p = 0 \Rightarrow 4r = 4 \Rightarrow r = 1$

Substituting $r = 1$ into the limit expression:

$\lim\limits_{x \to 2} \dfrac{x^2 + (p-2)x - 2p}{x-2} = 5$

$\lim\limits_{x \to 2} \dfrac{(x-2)(x+p)}{x-2} = 5$

$\lim\limits_{x \to 2} (x+p) = 5 \Rightarrow 2 + p = 5 \Rightarrow p = 3$

The quadratic equation is $rx^2 - px + q = 0$, which becomes $x^2 - 3x + q = 0$.

For both roots of $f(x) = x^2 - 3x + q = 0$ to lie in the interval $(0, 2)$, the following conditions must hold:

1) Discriminant $D \ge 0 \Rightarrow (-3)^2 - 4(1)(q) \ge 0 \Rightarrow 9 - 4q \ge 0 \Rightarrow q \le \dfrac{9}{4}$

2) $f(0) > 0 \Rightarrow q > 0$

3) $f(2) > 0 \Rightarrow 2^2 - 3(2) + q > 0 \Rightarrow q - 2 > 0 \Rightarrow q > 2$

4) The $x$-coordinate of the vertex must lie in $(0, 2) \Rightarrow 0 < \dfrac{-(-3)}{2(1)} < 2 \Rightarrow 0 < 1.5 < 2$, which is true.

Taking the intersection of all conditions, we get $q \in (2, \dfrac{9}{4}]$.

Comparing this with $(\alpha, \beta]$, we have $\alpha = 2$ and $\beta = \dfrac{9}{4}$.

Therefore, $4(\alpha + \beta) = 4(2 + \dfrac{9}{4}) = 8 + 9 = 17$.

Answer: $17$