Let $x$ be a positive real number such that $x + \frac{1}{x} = 98.$  Find\[\sqrt{x} + \frac{1}{\sqrt{x}}.\]
If I let $a = \sqrt{x}$ and $b = 1/\sqrt{x}$, then I get $(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = x + 2 + \frac{1}{x}.$ This looks promising, because I already know the value of $x + \frac{1}{x}.$ So I can substitute it and simplify: $(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = 98 + 2 = 100.$ Taking the square root of both sides, I get $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{100} = 10.$ 