[Number Theory]
Problem: Determine the two pairs of positive integers $(a, b)$ with $a<b$ that satisfy the equation $\sqrt{a}+\sqrt{b}=\sqrt{50}$.

Solution: First, we note that $\sqrt{50}=5 \sqrt{2}$.

Next, we note that $\sqrt{2}+4 \sqrt{2}=5 \sqrt{2}$ and $2 \sqrt{2}+3 \sqrt{2}=5 \sqrt{2}$.

From the first of these, we obtain $\sqrt{2}+\sqrt{32}=\sqrt{50}$.

From the second of these, we obtain $\sqrt{8}+\sqrt{18}=\sqrt{50}$.

Thus, $(a, b)=(2,32)$ and $(a, b)=(8,18)$ are solutions to the original equation.

(We are not asked to justify why these are the only two solutions.)

Final answer: (2,32), (8,18)