The question:

Given that,

$\displaystyle f(x) = x + \frac{1}{\sqrt{x}}$

$\displaystyle g(x) = \sqrt{x} - 2\sqrt{x}$

Find 6f + 3g.

My attempt:

$\displaystyle 6(x + \frac{1}{\sqrt{x}}) + 3(\sqrt{x} - 2\sqrt{x})$

$\displaystyle 6x + \frac{6}{\sqrt{x}} + 3\sqrt{x} - 6\sqrt{x}$

Apparently the answer is:

$\displaystyle 6x + 3\sqrt{x}$

I'm not sure how they were able to simplify the expression that way. I'm sure it's something obvious that I'm overlooking. Any assistance would be great.