Rationalizing Denominators: Practice

Multiply by the conjugate $\sqrt{5} - 1$. The denominator is $5 - 1 = 4$ and the numerator is $4(\sqrt{5} - 1)$.

$\frac{4}{\sqrt{5} + 1} = \frac{4(\sqrt{5} - 1)}{5 - 1} = \frac{4(\sqrt{5} - 1)}{4} = \sqrt{5} - 1$

The $4$ out front cancels the $4$ in the denominator.

Multiply by the conjugate $\sqrt{2} - 1$. The denominator is $(\sqrt{2})^2 - 1^2 = 2 - 1 = 1$, and the numerator is $\sqrt{2}(\sqrt{2} - 1) = 2 - \sqrt{2}$.

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

In the numerator $\sqrt{2}\cdot\sqrt{2} = 2$.

Multiply by the conjugate $\sqrt{7} + 3$. The denominator is $(\sqrt{7})^2 - 3^2 = 7 - 9 = -2$.

$\frac{1}{\sqrt{7} - 3} = \frac{\sqrt{7} + 3}{7 - 9} = \frac{\sqrt{7} + 3}{-2} = -\frac{\sqrt{7} + 3}{2}$

The negative denominator makes the whole result negative.

Multiply by the conjugate $\sqrt{5} + 1$. The denominator is $5 - 1 = 4$ and the numerator is $2(\sqrt{5} + 1)$.

$\frac{2}{\sqrt{5} - 1} = \frac{2(\sqrt{5} + 1)}{5 - 1} = \frac{2(\sqrt{5} + 1)}{4} = \frac{\sqrt{5} + 1}{2}$

The $2$ and $4$ reduce to $1$ and $2$.

Multiply by the conjugate $\sqrt{6} - \sqrt{3}$. The denominator is $6 - 3 = 3$ and the numerator is $3(\sqrt{6} - \sqrt{3})$.

$\frac{3}{\sqrt{6} + \sqrt{3}} = \frac{3(\sqrt{6} - \sqrt{3})}{6 - 3} = \frac{3(\sqrt{6} - \sqrt{3})}{3} = \sqrt{6} - \sqrt{3}$

The $3$ cancels.

Multiply by the conjugate $\sqrt{5} - \sqrt{2}$. The denominator is $5 - 2 = 3$.

$\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}$

The difference of squares gives $3$, not $7$.

Multiply top and bottom by the conjugate $1 + \sqrt{3}$. The denominator is $1^2 - (\sqrt{3})^2 = 1 - 3 = -2$, and the numerator is $(1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3}$.

$\frac{1 + \sqrt{3}}{1 - \sqrt{3}} = \frac{4 + 2\sqrt{3}}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$

Divide each term of the numerator by $-2$.

Multiply by the conjugate $\sqrt{6} - 1$. The denominator is $(\sqrt{6})^2 - 1^2 = 6 - 1 = 5$ and the numerator is $5(\sqrt{6} - 1)$.

$\frac{5}{\sqrt{6} + 1} = \frac{5(\sqrt{6} - 1)}{6 - 1} = \frac{5(\sqrt{6} - 1)}{5} = \sqrt{6} - 1$

The $5$ out front cancels the $5$ in the denominator.

Combine the denominator first: $\sqrt{8} = 2\sqrt{2}$, so $\sqrt{8} + \sqrt{2} = 3\sqrt{2}$.

$\frac{1}{\sqrt{8} + \sqrt{2}} = \frac{1}{3\sqrt{2}} = \frac{\sqrt{2}}{3\cdot 2} = \frac{\sqrt{2}}{6}$

Simplifying the radicals first turned the denominator into a single term.

Multiply by the conjugate $\sqrt{11} - 3$. The denominator is $(\sqrt{11})^2 - 3^2 = 11 - 9 = 2$.

$\frac{1}{\sqrt{11} + 3} = \frac{\sqrt{11} - 3}{11 - 9} = \frac{\sqrt{11} - 3}{2}$

The numerator takes the opposite sign, $\sqrt{11} - 3$.

Multiply by the conjugate $\sqrt{6} + \sqrt{2}$. The denominator is $6 - 2 = 4$ and the numerator is $8(\sqrt{6} + \sqrt{2})$.

$\frac{8}{\sqrt{6} - \sqrt{2}} = \frac{8(\sqrt{6} + \sqrt{2})}{6 - 2} = \frac{8(\sqrt{6} + \sqrt{2})}{4} = 2\sqrt{6} + 2\sqrt{2}$

The $8$ over $4$ reduces to $2$.

Multiply by the conjugate $\sqrt{3} - \sqrt{2}$. The denominator is $3 - 2 = 1$.

$\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$

Since the denominator is $1$, the answer is simply $\sqrt{3} - \sqrt{2}$.