If $N$ is a positive integer with $\sqrt{12}+\sqrt{108}=\sqrt{N}$, determine the value of $N$.
Since $\sqrt{12}+\sqrt{108}=\sqrt{N}$, then $\sqrt{2^{2} \cdot 3}+\sqrt{6^{2} \cdot 3}=\sqrt{N}$ or $2 \sqrt{3}+6 \sqrt{3}=\sqrt{N}$.

This means that $\sqrt{N}=8 \sqrt{3}=\sqrt{8^{2} \cdot 3}=\sqrt{192}$ and so $N=192$.

ANSWER: 192