Pre-Algebra Workbook 12: Radicals & Rationalizing the Denominator

To remove the square root from the denominator, multiply the numerator and denominator by $\sqrt{3}$

Multiply numerator and denominator by $\sqrt{3}$

$$\dfrac{5}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{5\sqrt{3}}{(\sqrt{3})(\sqrt{3})}$$

Since $(\sqrt{3})(\sqrt{3}) = 3$ the fraction becomes

$$\dfrac{5\sqrt{3}}{3}$$

The denominator is now rational.

The correct choice is B. $\dfrac{5\sqrt{3}}{3}$.

Multiply the top and bottom by $\sqrt{5}$ so that the denominator becomes $5$

Multiply numerator and denominator by $\sqrt{5}$

$$\dfrac{2}{\sqrt{5}} \cdot \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{2\sqrt{5}}{(\sqrt{5})(\sqrt{5})} = \dfrac{2\sqrt{5}}{5}$$

The denominator is now $5$ a rational number.

The correct choice is C. $\dfrac{2\sqrt{5}}{5}$.

Multiply numerator and denominator by $\sqrt{3}$ so the denominator becomes $3$ Use $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$

Multiply by $\dfrac{\sqrt{3}}{\sqrt{3}}$

$$\dfrac{\sqrt{2}}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{2}\sqrt{3}}{(\sqrt{3})(\sqrt{3})} = \dfrac{\sqrt{6}}{3}$$

The denominator is now the integer $3$

The correct choice is A. $\dfrac{\sqrt{6}}{3}$.

Treat $2\sqrt{3}$ as a single denominator. Multiply numerator and denominator by $\sqrt{3}$ then simplify the denominator.

Multiply by $\dfrac{\sqrt{3}}{\sqrt{3}}$

$$\dfrac{7}{2\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{7\sqrt{3}}{2(\sqrt{3}\cdot\sqrt{3})}$$

Compute $\sqrt{3} \cdot \sqrt{3} = 3$

$$\dfrac{7\sqrt{3}}{2 \cdot 3} = \dfrac{7\sqrt{3}}{6}$$

The denominator is now $6$

The correct choice is D. $\dfrac{7\sqrt{3}}{6}$.

You can combine the radicals first: $\dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{\dfrac{6}{2}}$ then simplify the fraction inside the radical.

Combine the square roots:

$$\dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{\dfrac{6}{2}} = \sqrt{3}$$

The result already has a rational denominator (since there is no denominator at all).

The correct choice is B. $\sqrt{3}$.

Multiply the entire numerator and the denominator by $\sqrt{5}$ Distribute $\sqrt{5}$ across $3 + \sqrt{5}$ in the numerator.

Multiply by $\dfrac{\sqrt{5}}{\sqrt{5}}$

$$\dfrac{3 + \sqrt{5}}{\sqrt{5}} \cdot \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{(3 + \sqrt{5})\sqrt{5}}{(\sqrt{5})(\sqrt{5})}$$

Distribute $\sqrt{5}$ in the numerator:

$$(3 + \sqrt{5})\sqrt{5} = 3\sqrt{5} + (\sqrt{5})(\sqrt{5}) = 3\sqrt{5} + 5$$

The denominator is

$$(\sqrt{5})(\sqrt{5}) = 5$$

So the fraction becomes

$$\dfrac{3\sqrt{5} + 5}{5}$$

The correct choice is A. $\dfrac{3\sqrt{5} + 5}{5}$.

Multiply numerator and denominator by $\sqrt{2}$ Distribute $\sqrt{2}$ over $4 - \sqrt{2}$ in the numerator, then simplify.

Multiply by $\dfrac{\sqrt{2}}{\sqrt{2}}$

$$\dfrac{4 - \sqrt{2}}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{(4 - \sqrt{2})\sqrt{2}}{(\sqrt{2})(\sqrt{2})}$$

Distribute $\sqrt{2}$

$$(4 - \sqrt{2})\sqrt{2} = 4\sqrt{2} - (\sqrt{2})(\sqrt{2}) = 4\sqrt{2} - 2$$

The denominator is $(\sqrt{2})(\sqrt{2}) = 2$ so

$$\dfrac{4\sqrt{2} - 2}{2}$$

Now simplify by dividing each term by $2$

$$\dfrac{4\sqrt{2}}{2} - \dfrac{2}{2} = 2\sqrt{2} - 1$$

The correct choice is C. $2\sqrt{2} - 1$.

First simplify $\sqrt{8}$ using a perfect square factor. Then see if any radicals cancel before worrying about rationalizing.

Simplify $\sqrt{8}$

$$8 = 4 \cdot 2, \quad \sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}$$

Substitute into the fraction:

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

Cancel $\sqrt{2}$ from numerator and denominator:

$$\dfrac{2}{4} = \dfrac{1}{2}$$

The simplified value is $\dfrac{1}{2}$ which already has a rational denominator.

The correct choice is A. $\dfrac{1}{2}$.

Notice the same radical appears in numerator and denominator. You can cancel $\sqrt{3}$ as long as it is not zero.

Write the fraction explicitly:

$$\dfrac{2\sqrt{3}}{\sqrt{3}}$$

Cancel $\sqrt{3}$ from top and bottom:

$$\dfrac{2\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}} = 2$$

The result is the integer $2$

The correct choice is D. $2$.

First simplify $\sqrt{8}$ then look for common radical factors to cancel. Remember $8 = 4\cdot 2$

Simplify $\sqrt{8}$

$$8 = 4 \cdot 2, \quad \sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}$$

Substitute into the fraction:

$$\dfrac{5\sqrt{2}}{\sqrt{8}} = \dfrac{5\sqrt{2}}{2\sqrt{2}}$$

Cancel $\sqrt{2}$ from numerator and denominator:

$$\dfrac{5}{2}$$

The simplified value is $\dfrac{5}{2}$

The correct choice is B. $\dfrac{5}{2}$.