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:

To solve this equation, let's look for a way to simplify it using substitution.

Notice that the terms $$(x - 6\sqrt{x})$$ and the $$9$$ from the last constant term can be grouped to form a perfect square:

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

Substitute this back into the original equation to reorganize it:

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

Let $$t = |\sqrt{x} - 3|$$. Since $$t$$ represents an absolute value, we know that $$t \ge 0$$.

The equation now becomes a standard quadratic in terms of $$t$$:

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

We can factor this quadratic by splitting the middle term:

$$t^2 - 2t + \sqrt{3}t - 2\sqrt{3} = 0$$

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

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

This gives two possible values for $$t$$:

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

Since we established that $$t \ge 0$$, we reject $$-\sqrt{3}$$. Therefore, $$t = 2$$.

Substitute back $$t = |\sqrt{x} - 3|$$:

$$|\sqrt{x} - 3| = 2$$

This results in two distinct cases:

1. $$\sqrt{x} - 3 = 2 \implies \sqrt{x} = 5 \implies x = 25$$
2. $$\sqrt{x} - 3 = -2 \implies \sqrt{x} = 1 \implies x = 1$$

The roots of the equation are $$1$$ and $$25$$. We are given that $$\alpha$$ and $$\beta$$ are the roots with $$\alpha < \beta$$, which means:

• $$\alpha = 1$$
• $$\beta = 25$$

Finally, substitute these values into the expression we need to evaluate:

$$\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$$

$$= \sqrt{\frac{25}{1}} + \sqrt{(1)(25)}$$

$$= 5 + 5 = 10$$