Rationalizing Denominators: Additional Practice (Set 2)

Multiply by the conjugate $\sqrt{7} + \sqrt{5}$. The denominator is $7 - 5 = 2$ and the numerator is $6(\sqrt{7} + \sqrt{5})$.

$\frac{6}{\sqrt{7} - \sqrt{5}} = \frac{6(\sqrt{7} + \sqrt{5})}{7 - 5} = \frac{6(\sqrt{7} + \sqrt{5})}{2} = 3\sqrt{7} + 3\sqrt{5}$

The $6$ over $2$ reduces to $3$.

Simplify the radical first, since $\sqrt{27} = 3\sqrt{3}$.

$\frac{9}{\sqrt{27}} = \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$

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

Multiply by the conjugate $\sqrt{10} + 3$. The denominator is $(\sqrt{10})^2 - 3^2 = 10 - 9 = 1$.

$\frac{1}{\sqrt{10} - 3} = \frac{\sqrt{10} + 3}{10 - 9} = \sqrt{10} + 3$

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

Multiply by the conjugate $\sqrt{7} - \sqrt{2}$. The denominator is $7 - 2 = 5$, and the numerator is $\sqrt{7}(\sqrt{7} - \sqrt{2}) = 7 - \sqrt{14}$.

$\frac{\sqrt{7}}{\sqrt{7} + \sqrt{2}} = \frac{7 - \sqrt{14}}{7 - 2} = \frac{7 - \sqrt{14}}{5}$

Use $\sqrt{7}\cdot\sqrt{7} = 7$ and $\sqrt{7}\cdot\sqrt{2} = \sqrt{14}$.

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

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

Here $\sqrt{5}\cdot\sqrt{5} = 5$ and $\sqrt{5}\cdot 2 = 2\sqrt{5}$.

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

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

The difference of squares gives $2$, not $10$.

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

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

The difference of squares gives $2$, not $6$.

Multiply by the conjugate $\sqrt{7} + 1$. The denominator is $7 - 1 = 6$ and the numerator is $12(\sqrt{7} + 1)$.

$\frac{12}{\sqrt{7} - 1} = \frac{12(\sqrt{7} + 1)}{7 - 1} = \frac{12(\sqrt{7} + 1)}{6} = 2\sqrt{7} + 2$

The $12$ over $6$ reduces to $2$.

To clear a two-term denominator, multiply by the conjugate over itself. The conjugate of $3 - \sqrt{2}$ is $3 + \sqrt{2}$.

$(3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$

Multiplying by $\frac{3 - \sqrt{2}}{3 - \sqrt{2}}$ would give $(3 - \sqrt{2})^2 = 11 - 6\sqrt{2}$, which still has a root.

Multiply top and bottom by the conjugate $2 + \sqrt{3}$. The denominator is $2^2 - (\sqrt{3})^2 = 4 - 3 = 1$, and the numerator is $(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$.

$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{7 + 4\sqrt{3}}{4 - 3} = 7 + 4\sqrt{3}$

Expand $(2 + \sqrt{3})^2$ as a square of a binomial.

Multiply top and bottom by the conjugate $\sqrt{5} + 1$. The denominator is $(\sqrt{5})^2 - 1^2 = 5 - 1 = 4$, and the numerator is $(\sqrt{5} + 1)^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$.

$\frac{\sqrt{5} + 1}{\sqrt{5} - 1} = \frac{6 + 2\sqrt{5}}{5 - 1} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$

Divide the numerator and denominator by $2$.

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

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

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