The Conjugate Below

A bottom like $4-\sqrt{6}$ is harder. Multiplying it by $\sqrt{6}$ gives $4\sqrt{6}-6$, so the root only moves to the other term. You need the pair from the last chapter, the conjugate: the same two terms with the sign between them flipped. The conjugate of $4-\sqrt{6}$ is $4+\sqrt{6}$.

A sum times its difference loses its middle terms: $\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right) = {a}^{2}-b$. When a and b are whole numbers, so is that, and the root is gone. Take $\frac{10}{4-\sqrt{6}}$ and multiply the top and the bottom by $4+\sqrt{6}$:

$$\frac{10}{4-\sqrt{6}} = \frac{10\left(4+\sqrt{6}\right)}{16-6} = \frac{10\left(4+\sqrt{6}\right)}{10}$$

The 10s cancel and the answer is $4+\sqrt{6}$. Keep the top in its bracket until the bottom is worked out: the bottom is a whole number that often shares a factor with the number in front, as the 10s did here, and multiplying out early would hide that.