simplifying the quotient of two radical expressions

Question

How do we find $p$ and $q$ so that $$\frac{\sqrt{2}+2}{\sqrt{2}-1}=p+q\sqrt{2}\,?$$

Answer 1

Just multiply numerator and denominator by the "conjugate" $\sqrt{2}+1$, giving $$\frac{\sqrt{2}+2}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{1}=4+3\sqrt{2}$$