Let $$a > 0$$ be a root of the equation $$2x^2 + x - 2 = 0$$. If $$\lim_{x \to \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = \alpha + \beta\sqrt{17}$$, where $$\alpha, \beta \in Z$$, then $$\alpha + \beta$$ is equal to ______

We need to find $$\alpha + \beta$$ where $$\lim_{x \to 1/a} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = \alpha + \beta\sqrt{17}$$.

Find $$a$$. From $$2x^2 + x - 2 = 0$$ with $$a > 0$$:

$$ a = \frac{-1 + \sqrt{1 + 16}}{4} = \frac{-1 + \sqrt{17}}{4} $$

As $$x \to \frac{1}{a}$$, we need $$2 + x - 2x^2 \to 0$$ (since $$\frac{1}{a}$$ is a root of $$2x^2 - x - 2 = 0$$, i.e., $$2x^2 = x + 2$$, so $$2 + x - 2x^2 = 2 + x - (x+2) = 0$$). ✓

So both numerator and denominator approach 0. Use the limit $$\lim_{u \to 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$$.

$$ \lim_{x \to 1/a} \frac{16(1 - \cos(2 + x - 2x^2))}{(1 - ax)^2} = 16 \cdot \frac{1}{2} \cdot \lim_{x \to 1/a} \frac{(2 + x - 2x^2)^2}{(1 - ax)^2} $$

$$ = 8 \cdot \left(\lim_{x \to 1/a} \frac{2 + x - 2x^2}{1 - ax}\right)^2 $$

Evaluate the limit. Factor $$2 + x - 2x^2 = -2(x^2 - x/2 - 1) = -2(x - 1/a)(x + 1/(2\cdot something))$$.

$$2 + x - 2x^2 = -(2x^2 - x - 2)$$. The roots of $$2x^2 - x - 2 = 0$$ are $$x = \frac{1 \pm \sqrt{17}}{4}$$.

So $$2 + x - 2x^2 = -2\left(x - \frac{1+\sqrt{17}}{4}\right)\left(x - \frac{1-\sqrt{17}}{4}\right)$$

And $$\frac{1}{a} = \frac{4}{-1+\sqrt{17}} = \frac{4(\sqrt{17}+1)}{16} = \frac{\sqrt{17}+1}{4} = \frac{1+\sqrt{17}}{4}$$.

Also $$1 - ax = -a(x - 1/a)$$.

$$ \frac{2 + x - 2x^2}{1 - ax} = \frac{-2(x - \frac{1+\sqrt{17}}{4})(x - \frac{1-\sqrt{17}}{4})}{-a(x - \frac{1+\sqrt{17}}{4})} = \frac{2}{a}\left(x - \frac{1-\sqrt{17}}{4}\right) $$

At $$x = \frac{1+\sqrt{17}}{4}$$:

$$ = \frac{2}{a} \cdot \left(\frac{1+\sqrt{17}}{4} - \frac{1-\sqrt{17}}{4}\right) = \frac{2}{a} \cdot \frac{2\sqrt{17}}{4} = \frac{\sqrt{17}}{a} $$

With $$a = \frac{-1+\sqrt{17}}{4}$$:

$$ = \frac{\sqrt{17}}{\frac{\sqrt{17}-1}{4}} = \frac{4\sqrt{17}}{\sqrt{17}-1} = \frac{4\sqrt{17}(\sqrt{17}+1)}{16} = \frac{\sqrt{17}(\sqrt{17}+1)}{4} = \frac{17+\sqrt{17}}{4} $$

$$ \text{Limit} = 8 \cdot \left(\frac{17+\sqrt{17}}{4}\right)^2 = 8 \cdot \frac{(17+\sqrt{17})^2}{16} = \frac{(17+\sqrt{17})^2}{2} $$

$$ = \frac{289 + 34\sqrt{17} + 17}{2} = \frac{306 + 34\sqrt{17}}{2} = 153 + 17\sqrt{17} $$

So $$\alpha = 153$$, $$\beta = 17$$, and $$\alpha + \beta = 170$$.

The answer is 170.