Question

$\ (x - 6\sqrt{x} + 9) - (2 - \sqrt{3})|\sqrt{x} - 3| - 2\sqrt{3} = 0$

$$\Rightarrow |\sqrt{x} - 3|^{2} - (2 - \sqrt{3})|\sqrt{x} - 3| - 2\sqrt{3} = 0 $$$\Rightarrow |\sqrt{x} - 3| = 2$ or $|\sqrt{x} - 3| = - \sqrt{3}$ (not possible)

$\Rightarrow \sqrt{x} = 1$ or 5

$\Rightarrow x = 1$ or 5

$\Rightarrow \alpha = 1$ and $\beta = 25$

Alter:

Let $x \geq 9$, let $\sqrt{x} = t \Rightarrow t \geq 3$

$$(\sqrt{3} - 2)(t - 3) + (t - 3)^{2} - 2\sqrt{3} = 0 $$Let $t - 3 = u$

$$u^{2} + (\sqrt{3} - 2)u - 2\sqrt{3} = 0 $$$u = 2$, or $u = - \sqrt{3}$

$\Rightarrow t - 3 = 2$ or $t - 3 = - \sqrt{3}$

$\Rightarrow t = 5$ or $t = 3 - \sqrt{3}\ $ (rejected)

$$\Rightarrow x = 25 $$Now let $0 < x < 9$,

$$- (\sqrt{3} - 2)(t - 3) + (t - 3)^{2} - 2\sqrt{3} = 0 $$let $t - 3 = u$

$$u^{2} - (\sqrt{3} - 2)u - 2\sqrt{3} = 0 $$$\Rightarrow u = \sqrt{3}$ or $u = - 2$

$\Rightarrow t = 3 + \sqrt{3}$ (rejected) or $t - 3 = - 2$

$${t = 1 \Rightarrow x = 1 }{\alpha = 1,\beta = 25 }$$Now $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 10$