Radicals and Fractions

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Algebra II › Radicals and Fractions

Questions 1 - 10

1

Simplify, and ensure that no radicals remain in the denominator.

$\frac{1}{\sqrt{252}}$

$\frac{\sqrt{7}}{42}$

None of these

$\frac{\sqrt{7}}{252}$

$\frac{\sqrt{7}}{21}$

$\frac{\sqrt{7}}{36}$

Explanation

$\frac{1}{\sqrt{252}}$

Moving radical from the denominator to the numerator:

$\frac{1}{\sqrt{252}}*\frac{\sqrt{252}}{\sqrt{252}}=\frac{\sqrt{252}}{252}$

Factoring:

$\frac{\sqrt{252}}{252}=\frac{\sqrt{9}*\sqrt{4}*\sqrt{7}}{252}$

Simplifying:

$\frac{\sqrt{9}*\sqrt{4}*\sqrt{7}}{252}=\frac{\sqrt{7}}{42}$

2

Simplify the fraction: $\frac{21}{\sqrt{7}}$

$3\sqrt{7}$

$\frac{3\sqrt{7}}{7}$

$\frac{\sqrt{7}}{3}$

$\frac{\sqrt{21}}{3}$

$\frac{\sqrt{21}}{7}$

Explanation

Multiply the numerator and denominator by the denominator.

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

Reduce the fraction.

The answer is: $3\sqrt{7}$

3

Simplify:

$\sqrt{\frac{9}{16}}$

$\frac{3}{4}$

$\frac{\sqrt{9}}{8}$

$\frac{3}{16}$

$\frac{3}{8}$

Explanation

We can take the square roots of the numerator and denominator separately. Thus, we get:

$\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$

4

$\sqrt{\frac{25x^4}{225y^{10}}}$

$\frac{x^2}{3y^2}$

$\frac{4x^2}{3y^2}$

$\frac{x^2}{y^2}$

$\frac{x^2}{3y}$

$\frac{2x^2}{3y^2}$

Explanation

To simplify, first start by rewriting as:

$\frac{\sqrt{25x^4}}{\sqrt{225y^{10}}}$ .

Then, simplify the numerator and denominator separately.

$\sqrt{25x^4}=5x^2$

and

$\sqrt{225y^{10}}=15y^5$ .

Now, you have

$\frac{5x^2}{15y^5}$ .

Simplify to get

$\frac{x^2}{3y^2}$ .

5

Rationalize the denominator. $\frac{\sqrt{6}}{\sqrt5}$

$\frac{\sqrt{30}}{5}$

$\frac{3\sqrt{10}}{5}$

$\frac{\sqrt{30}}{2}$

$\frac{2\sqrt{15}}{5}$

$\frac{6\sqrt{5}}{5}$

Explanation

In order to rationalize the denominator, multiply both the numerator and denominator by square root five.

$\frac{\sqrt{6}}{\sqrt5} \cdot \frac{\sqrt5}{\sqrt5}$

When a radical of a certain number is multiplied by itself, the radical will be eliminated, leaving only the integer.

$\frac{\sqrt{6}}{\sqrt5} \cdot \frac{\sqrt5}{\sqrt5} = \frac{\sqrt{30}}{5}$

This cannot be simplified any further.

The answer is: $\frac{\sqrt{30}}{5}$

6

Simplify: $\frac{\sqrt{12}}{2\sqrt{24}}$

$\frac{\sqrt2}{4}$

$\sqrt{3}$

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

$\frac{4\sqrt{6}}{3}$

$\frac{5\sqrt2}{4}$

Explanation

Simplify both the top and bottom of the fraction.

$\frac{\sqrt{12}}{2\sqrt{24}}=\frac{\sqrt{4}\cdot \sqrt3}{2\sqrt{4}\cdot \sqrt{6}}$

Cancel the common terms.

$\frac{\sqrt3}{2\sqrt6}$

Rationalize the denominator by multiplying the top and bottom by root six.

$\frac{\sqrt3}{2\sqrt6}\cdot \frac{\sqrt6}{\sqrt6}$

Multiply the numerator with numerator and denominator with denominator.

$\frac{\sqrt{18}}{2 \times 6} = \frac{\sqrt9\cdot \sqrt2}{12} = \frac{3\sqrt2}{12}$

Simplify the fraction.

The answer is: $\frac{\sqrt2}{4}$

7

Rationalize the denominator: $\frac{6}{\sqrt{30}}$

$\frac{\sqrt{30}}{5}$

$\frac{\sqrt{30}}{6}$

$\frac{\sqrt{30}}{3}$

$\sqrt{10}$

$\frac{2\sqrt{10}}{3}$

Explanation

In order to eliminate the radical sign, multiply the top and the bottom by the denominator.

$\frac{6}{\sqrt{30}}\cdot \frac{\sqrt{30}}{\sqrt{30}} = \frac{6\sqrt{30}}{30}$

Reduce the numerator and denominator.

$\frac{6\sqrt{30}}{6\times 5}= \frac{\sqrt{30}}{5}$

The answer is: $\frac{\sqrt{30}}{5}$

8

Simplify, if possible: $\frac{9}{\sqrt[3]9}$

$3\sqrt[3]{3}$

$\textup{The expression is simplified as is.}$

$\frac{3\sqrt[3]{3}}{2}$

$4\sqrt[3]{3}$

$6\sqrt[3]{3}$

Explanation

In order to eliminate the radical in the denominator, we will need to multiply the denominator twice on both the numerator and denominator.

$\frac{9}{\sqrt[3]9}\cdot \frac{\sqrt[3]9}{\sqrt[3]9} \cdot \frac{\sqrt[3]9}{\sqrt[3]9} = \frac{9\sqrt[3]9\cdot \sqrt[3]9}{9}$

Simplify the fraction.

$\sqrt[3]9\cdot \sqrt[3]9 = \sqrt[3]{81}$

Factor the inner term using common factors of numbers the power of three.

$\sqrt[3]{81}=\sqrt[3]{27\times 3} = \sqrt[3]{27}\cdot\sqrt[3]{3} = 3\sqrt[3]{3}$

The answer is: $3\sqrt[3]{3}$

9

Rationalize the denominator: $\frac{2}{2+\sqrt{3}}$

$4-2\sqrt3$

$2-3\sqrt3$

$2-4\sqrt3$

$2-4\sqrt3$

Explanation

Multiply the top and bottom by the conjugate of the denominator.

$\frac{2}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$

Simplify the top and bottom of the fractions.

$\frac{2}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} =\frac{4-2\sqrt3}{4-3}=4-2\sqrt3$

The answer is: $4-2\sqrt3$

10

Evaluate: $\frac{5}{\sqrt{3}\cdot \sqrt{5}}$

$\frac{ \sqrt{15}}{3}$

$\frac{ \sqrt{5}}{3}$

$\frac{ \sqrt{15}}{5}$

$\frac{ 3\sqrt{5}}{5}$

$\frac{ 5\sqrt{5}}{3}$

Explanation

Multiply the denominator together to combine as one radical.

$\frac{5}{\sqrt{3}\cdot \sqrt{5}} = \frac{5}{\sqrt{15}}$

Rationalize the denominator by multiplying the top and bottom of the fraction by the denominator.

$\frac{5}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}} =\frac{ 5\sqrt{15}}{15}$

Reduce the fraction.

The answer is: $\frac{ \sqrt{15}}{3}$

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