If $$x = 5 + 2\sqrt{6}$$, find the value of $$\frac{x - 1}{\sqrt{x}}$$.

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Accepted Answer

Given $$x = 5 + 2\sqrt{6}$$, rewrite it as $$x = 3 + 2 + 2\sqrt{6} = (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \cdot \sqrt{3} \cdot \sqrt{2} = (\sqrt{3} + \sqrt{2})^2$$. Thus, $$x = (\sqrt{3} + \sqrt{2})^2$$.

Now, calculate $$x - 1$$:  
$$
x - 1 = (5 + 2\sqrt{6}) - 1 = 4 + 2\sqrt{6} = 2(2 + \sqrt{6}) = 2\sqrt{2}(\sqrt{3} + \sqrt{2}).
$$

Next, compute $$\frac{x - 1}{\sqrt{x}}$$:  
Since $$x = (\sqrt{3} + \sqrt{2})^2$$, we have $$\sqrt{x} = \sqrt{3} + \sqrt{2}$$. Therefore:  
$$
\frac{x - 1}{\sqrt{x}} = \frac{2\sqrt{2}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}} = 2\sqrt{2}.
$$

Thus, the answer is **$$b) 2\sqrt{2}$$**.

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