Rationalizing Denominators: Practice

Multiply the top and bottom by $\sqrt{2}$ so the denominator $\sqrt{2}\cdot\sqrt{2}$ becomes $2$.

$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

The root moves to the numerator; it does not disappear, so $\frac{1}{2}$ is wrong.

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

$\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Multiplying the denominator by itself gives $3$, not $9$, so $\frac{\sqrt{3}}{9}$ is wrong.

A square root times itself gives back the number under the root.

$\sqrt{7}\cdot\sqrt{7} = \sqrt{7\cdot 7} = \sqrt{49} = 7$

The answer is $7$, not $49$ (that would be $7^2$) and not $14$.

Multiply the top and bottom by $\sqrt{3}$. The denominator becomes $3$ and the $2$ stays out front.

$\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Dropping the $2$ gives $\frac{\sqrt{3}}{3}$, a different value.

Multiplying by $\frac{\sqrt{b}}{\sqrt{b}}$ (a form of $1$) turns the denominator $\sqrt{b}$ into $\sqrt{b}\cdot\sqrt{b} = b$.

$\frac{a}{\sqrt{b}}\cdot\frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$

So you multiply the top and bottom by $\sqrt{b}$.

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

$\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

The root moves to the numerator, so $\frac{1}{7}$ is wrong.

The conjugate flips the sign between the two terms.

$5 - \sqrt{2}\ \longrightarrow\ 5 + \sqrt{2}$

Multiplying the two gives $5^2 - (\sqrt{2})^2 = 25 - 2 = 23$, a rational number.

Any nonzero quantity divided by itself equals $1$.

$\frac{\sqrt{3}}{\sqrt{3}} = 1$

Multiplying by this form of $1$ changes the look of a fraction but not its value, which is why rationalizing is allowed.

Multiply the top and bottom by $\sqrt{7}$. The denominator becomes $7$ and the $2$ stays out front.

$\frac{2}{\sqrt{7}} = \frac{2}{\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{7}}{7}$

Dropping the $2$ gives $\frac{\sqrt{7}}{7}$, a different value.

The conjugate flips the sign between the two roots.

$\sqrt{5} + \sqrt{2}\ \longrightarrow\ \sqrt{5} - \sqrt{2}$

Their product is $(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$.

Multiply the top and bottom by $\sqrt{2}$. The denominator becomes $2$ and the $3$ stays out front.

$\frac{3}{\sqrt{2}} = \frac{3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

The root does not cancel, so $\frac{3}{2}$ is wrong.

Multiply the top and bottom by $\sqrt{5}$, then reduce the fraction.

$\frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}$

The $5$ on top cancels the $5$ on the bottom, leaving $\sqrt{5}$.