Rationalizing Denominators Using Conjugates

For expressions like $\frac{1}{3-\sqrt{2}}$, if we wish to rationalize the denominator, it is not enough to simply multiply the numerator and denominator by $\sqrt{2}$ as that would not get rid of all radicals in the denominator.

However, we can use a similar trick, taking advantage of differences of squares. Notice

$$\frac{1}{3-\sqrt{2}}\cdot\frac{3+\sqrt{2}}{3+\sqrt{2}}=\frac{3+\sqrt{2}}{3^2-(\sqrt{2})^2}$$

And the radical gets cancelled with the square, giving you $\frac{3+\sqrt{2}}{9-2}=\frac{3+\sqrt{2}}{7}.$

The method here is to multiply the numerator and denominator by the conjugate of the denominator; i.e. the same denominator, but with the opposite sign between the two terms you see.
- E.g. for $\frac{1}{1+\sqrt{5}}$ multiply top and bottom by $1-\sqrt{5}$
- E.g. for $\frac{1}{1-\sqrt{5}}$ multiply top and bottom by $1+\sqrt{5}$
Multiplying by the conjugate of the denominator gives cancellation of all radicals via difference of squares.

Solution: $-3+\sqrt{10}$

Solution: $-3\sqrt{2}+3\sqrt{3}$

Solution: $-2\sqrt{2}-2\sqrt{3}$ or $-2(\sqrt{2}+\sqrt{3})$

Solution: $\frac{\sqrt{5}+3\sqrt{7}-\sqrt{35}-21}{-58}$

Solution: $\frac{9|x|-3\sqrt{xy}}{9|x|-|y|}$