Rationalizing Denominators: Practice

Multiply by $\frac{\sqrt{2}}{\sqrt{2}}$, then reduce.

$\frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$

The $8$ and the $2$ share a factor of $2$, so the fraction simplifies to $4\sqrt{2}$.

Multiply the top and bottom by $\sqrt{5}$. The top becomes $\sqrt{2}\cdot\sqrt{5} = \sqrt{10}$ and the bottom becomes $5$.

$\frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{2}\cdot\sqrt{5}}{5} = \frac{\sqrt{10}}{5}$

The roots combine under one radical since $\sqrt{2}\cdot\sqrt{5} = \sqrt{10}$.

Multiply by the conjugate $2 + \sqrt{3}$. The denominator is $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$.

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

Since the difference of squares is $1$, the denominator disappears.

Multiply by the conjugate $3 - \sqrt{5}$. The denominator is $3^2 - (\sqrt{5})^2 = 9 - 5 = 4$.

$\frac{1}{3 + \sqrt{5}} = \frac{3 - \sqrt{5}}{9 - 5} = \frac{3 - \sqrt{5}}{4}$

The numerator takes the opposite sign, $3 - \sqrt{5}$.

Simplify the radical first, since $\sqrt{12} = 2\sqrt{3}$.

$\frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$

After cancelling, $\frac{3}{\sqrt{3}}$ rationalizes to $\sqrt{3}$.

The conjugate flips the sign between the two terms.

$\sqrt{7} - 2\ \longrightarrow\ \sqrt{7} + 2$

Multiplying the two gives $(\sqrt{7})^2 - 2^2 = 7 - 4 = 3$, a rational number.

Simplify the radical first, since $\sqrt{8} = 2\sqrt{2}$.

$\frac{4}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

The fraction reduces all the way down to $\sqrt{2}$.

Multiply by the conjugate $\sqrt{7} + \sqrt{5}$. The denominator is $7 - 5 = 2$.

$\frac{1}{\sqrt{7} - \sqrt{5}} = \frac{\sqrt{7} + \sqrt{5}}{7 - 5} = \frac{\sqrt{7} + \sqrt{5}}{2}$

The difference of squares gives $2$, not $12$; adding the numbers is the common slip.

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

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

Dividing by $-1$ flips the sign of each term.

Multiply the top and bottom by $\sqrt{7}$. The top becomes $\sqrt{3}\cdot\sqrt{7} = \sqrt{21}$ and the bottom becomes $7$.

$\frac{\sqrt{3}}{\sqrt{7}} = \frac{\sqrt{3}\cdot\sqrt{7}}{7} = \frac{\sqrt{21}}{7}$

The roots combine as $\sqrt{3}\cdot\sqrt{7} = \sqrt{21}$.

Multiply by the conjugate $\sqrt{6} + \sqrt{2}$. The denominator is $6 - 2 = 4$.

$\frac{1}{\sqrt{6} - \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{6 - 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

The difference of squares gives $4$, not $8$.

Multiply by $\frac{\sqrt{6}}{\sqrt{6}}$, then reduce.

$\frac{12}{\sqrt{6}} = \frac{12\sqrt{6}}{6} = 2\sqrt{6}$

The $12$ over $6$ reduces to $2$, leaving $2\sqrt{6}$.