Root Times Root

The rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$ also runs the other way: for numbers that are not negative, two roots multiplied make one root of the product. So $\sqrt{2}\cdot \sqrt{8}$ is $\sqrt{16}$, which is 4. Neither root was a whole number, yet their product is.

The new root may need simplifying. $\sqrt{6}\cdot \sqrt{10}$ is $\sqrt{60}$, and 60 holds the square 4:

$$\sqrt{6}\cdot \sqrt{10} = \sqrt{60} = \sqrt{4}\cdot \sqrt{15} = 2\sqrt{15}$$

Numbers in front multiply with each other, roots with roots: $\left(2\sqrt{3}\right)\left(4\sqrt{5}\right)$ is 8 times $\sqrt{15}$. And a root times itself is what was under it: $\sqrt{7}\cdot \sqrt{7}$ is 7.