If $$\alpha$$ and $$\beta$$ ($$\alpha < \beta$$) are the roots of the equation $$(-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6\sqrt{x})+(9-2\sqrt{3})=0,x\geq0\text{ then }\sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha\beta}$$ is equal to:

 $$t = \sqrt{x}$$ so that $$x = t^2$$ and $$t \ge 0$$. Substituting into the given equation we get

$$(-2 + \sqrt{3})\,|t - 3| + (t^2 - 6t) + (9 - 2\sqrt{3}) = 0\,. $$

We will consider two cases depending on the sign of $$t - 3$$.

Case 1: $$t \ge 3\,. $$

Then $$|t - 3| = t - 3$$ and the equation becomes

$$(-2 + \sqrt{3})(t - 3) + t^2 - 6t + 9 - 2\sqrt{3} = 0\,. $$

$$(-2 + \sqrt{3})t -(-2 + \sqrt{3})\cdot 3 + t^2 - 6t + 9 - 2\sqrt{3} = 0$$

$$\Rightarrow t^2 + (-8 + \sqrt{3})\,t + (15 - 5\sqrt{3}) = 0\quad\,-(1)$$

$$t = \frac{8 - \sqrt{3} + \sqrt{3} + 2}{2} = 5\quad\text{or}\quad t = \frac{8 - \sqrt{3} - \sqrt{3} - 2}{2} = 3 - \sqrt{3}\,. $$

Since we are in Case 1 with $$t \ge 3$$, only $$t = 5$$ is valid. This gives

$$x = t^2 = 25\,. $$

Case 2: $$0 \le t < 3\,. $$

Then $$|t - 3| = 3 - t$$ and the equation becomes

$$(-2 + \sqrt{3})(3 - t) + t^2 - 6t + 9 - 2\sqrt{3} = 0\,. $$

$$(-2 + \sqrt{3})\cdot 3 -(-2 + \sqrt{3})t + t^2 - 6t + 9 - 2\sqrt{3} = 0$$

$$\Rightarrow t^2 + (-4 - \sqrt{3})\,t + (3 + \sqrt{3}) = 0\quad\,-(2)$$

$$t = \frac{4 + \sqrt{3} + \sqrt{3} + 2}{2} = 3 + \sqrt{3}\quad\text{or}\quad t = \frac{4 + \sqrt{3} - \sqrt{3} - 2}{2} = 1\,. $$

In Case 2 with $$t < 3$$, only $$t = 1$$ is valid. This gives

$$x = t^2 = 1\,. $$

Therefore the two roots of the original equation (with $$\alpha < \beta$$) are

$$\alpha = 1\,,\quad \beta = 25\,. $$

$$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\,\beta} = \sqrt{\frac{25}{1}} + \sqrt{1 \cdot 25} = 5 + 5 = 10\,. $$