If alpha and beta (alpha

Question

(C) The given equation is $(\sqrt{3}-2)|\sqrt{x}-3| + (\sqrt{x}-3)^2 - 2\sqrt{3} = 0$. Let $t = |\sqrt{x}-3|$. Then the equation becomes $t^2 + (\sqrt

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The given equation is $(\sqrt{3}-2)|\sqrt{x}-3| + (\sqrt{x}-3)^2 - 2\sqrt{3} = 0$. Let $t = |\sqrt{x}-3|$. Then the equation becomes $t^2 + (\sqrt{3}-2)t - 2\sqrt{3} = 0$. Factoring the quadratic: $(t+ \sqrt{3})(t-2) = 0$. Since $t = |\sqrt{x}-3| \ge 0$, we must have $t = 2$. Thus, $|\sqrt{x}-3| = 2$, which implies $\sqrt{x}-3 = 2$ or $\sqrt{x}-3 = -2$. This gives $\sqrt{x} = 5$ or $\sqrt{x} = 1$. So, $x = 25$ or $x = 1$. Given $\alpha Finally, $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta} = \sqrt{\frac{25}{1}} + \sqrt{1 	imes 25} = 5 + 5 = 10$.

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

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