Radicals and Fractions - Algebra 2

Card 1 of 212

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Simplify the following expression:

$\frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}$

Tap to reveal answer

Answer

First, let's see how we can combine these two fractions. Remember the following relationships:

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

and

$\sqrt{a}\sqrt{b}=\sqrt{ab}$

Now, let's look at our problem. Let's first try and turn the first term into one big radical:

$\frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}$

$\frac{\sqrt{27}}{\sqrt{4}}\cdot \sqrt{\frac{1}{108}}$

$\sqrt{\frac{27}{4}}\cdot \sqrt{\frac{1}{108}}$

Great! We've used the first relationship; now let's combine the two radicals using the second relationship.

$\sqrt{\frac{27}{4}\cdot\frac{1}{108}}=\sqrt{\frac{27}{4\cdot 108}}$

I haven't multiplied out anything yet because I want to see if there's any simplifying I can do BEFORE I multiply. In this case, I ask myself: Does the denominator contain any factors of 27 (3, 9, 27)?

I know 108 is divisible by 9 because its digits add up to a number that's divisible by 9. 27 is divisible by 9 too, so I can rewrite it this way:

$\sqrt{\frac{3\cdot \mathbf{9}}{4\cdot \mathbf{9}\cdot 12}}=\sqrt{\frac{3}{4\cdot 12}}$

We also know that 3 is a factor of 12:

$\sqrt{\frac{\mathbf{3}}{4\cdot 4\cdot \mathbf{3}}}=\sqrt{\frac{1}{4\cdot 4}}$

Now, after simplifying the fraction, we have to simplify the radical. Keep this in mind:

$\sqrt{a\cdot a}=\sqrt{a^{2}}=a$

We can finally simplify this expression completely:

$\sqrt{\frac{1}{4\cdot 4}}=\frac{\sqrt{1}}{\sqrt{4\cdot 4}}=\frac{1}{4}$

← Didn't Know|Knew It →

Question

Rationalize the denominator.

$\frac{x^{2}}{\sqrt{11}}$

Tap to reveal answer

Answer

In order to rationalize the denominator we must eliminate the root in the denominator.

To do this, we multiply the radical by $\frac{\sqrt{11}}{\sqrt{11}}$ ,

$\frac{x^{2}}{\sqrt{11}}\cdot \frac{\sqrt{11}}{\sqrt{11}}=\frac{x^{2}\sqrt{11}}{(\sqrt{11})^{2}}=\frac{x^{2}\sqrt{11}}{11}$

← Didn't Know|Knew It →

Question

Simplify by rationalizing the denominator:

$\frac{14}{8 - \sqrt{15}}$

Tap to reveal answer

Answer

Multiply the numerator and the denominator by the conjugate of the denominator, which is $8 + \sqrt{15}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{14}{8 - \sqrt{15}}$

$= \frac{14(8 + \sqrt{15})}{\left (8 - \sqrt{15} \right )(8 + \sqrt{15})}$

$= \frac{14(8 + \sqrt{15})}{8^{2} - \left ( \sqrt{15} \right )^{2} }$

$= \frac{14(8 + \sqrt{15})}{64- 15}$

$= \frac{14(8 + \sqrt{15})}{49}$

$= \frac{2(8 + \sqrt{15})}{7}$

$= \frac{16+2 \sqrt{15}}{7}$

← Didn't Know|Knew It →

Question

Simplify:

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

Tap to reveal answer

Answer

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

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

← Didn't Know|Knew It →

Question

Simplify $\frac{\sqrt 32}{\sqrt2}$ .

Tap to reveal answer

Answer

Understanding properties of radicals will help you quickly solve this problem. When two radicals are multiplied or divided, you can simply combine the two radicals. For instance: $\sqrt{x}\cdot \sqrt{y} = \sqrt {xy}$

For this equation:

$\sqrt{\frac{32}{2}} = \sqrt{16} = \pm4$

← Didn't Know|Knew It →

Question

Simplify the following:

$\frac{4\sqrt{5}}{\sqrt{20}}$

Tap to reveal answer

Answer

You can begin by rewriting this equation as:

$4 * \frac{\sqrt{5}}{\sqrt{20}}=4 * \sqrt{\frac{5}{20}} = 4 * \sqrt{\frac{1}{4}}=\frac{4}{\sqrt{4}}$

Now, you need to rationalize the denominator. To do this, multiply both top and bottom by $\sqrt{4}$ :

$\frac{4}{\sqrt{4}} * \frac{\sqrt{4}}{\sqrt{4}} = \frac{4\sqrt{4}}{4}$

Then, cancel the common :

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

Since is a perfect square you can take the square root to get the simplified answer.

$\sqrt4=\pm 2$

← Didn't Know|Knew It →

Question

Solve and simplify.

$(3+\sqrt{5})^2$

Tap to reveal answer

Answer

When foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables.

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

$=9+3\sqrt{5}+3\sqrt{5}+5$

Combining like terms we get our final answer as follows.

$=14+6\sqrt{5}$

← Didn't Know|Knew It →

Question

Solve and simplify.

$\frac{9}{4\sqrt{5}}$

Tap to reveal answer

Answer

When dividing radicals, check the denominator to make sure it can be simplified or that there is a radical present that needs to be fixed. Since there is a radical present, we need to eliminate that radical. To do this, we multiply both top and bottom by $\sqrt{5}$ . The reason is because we want a whole number in the denominator and multiplying by itself will achieve that. By multiplying itself, it creates a square number which can be reduced to .

With the denominator being $4\cdot 5$ , the numerator is $9\cdot \sqrt{5}=9\sqrt{5}$ .

Final answer is $\frac{9\sqrt{5}}{20}$ .

← Didn't Know|Knew It →

Question

Simplify $\sqrt{116x^{11}y^4z^7}$

Tap to reveal answer

Answer

To simplify radicals, I like to approach each term separately. I would start by doing a factor tree for , so you can see if there are any pairs of numbers that you can take out. factors to $29\cdot 2\cdot 2$ , so you can take a out of the radical. For $x^{11}$ , there are pairs of 's, so $x^{5}$ goes outside of the radical, and one remains underneath the radical. For , there are pairs of 's, so you can take 's outside the radical. For , there are complete pairs of 's so goes on the outside, while one remains underneath the radical. Now, put those all together to get: $2x^5y^2z^3\sqrt{29xz}$ .

← Didn't Know|Knew It →

Question

Simplify $\frac{\sqrt{64x^8y^{12}z^{10}}}{20x^{-6}y^4z^2}$

Tap to reveal answer

Answer

To simplify this expression, I would start by simplifying the radical on the numerator. Remember, for every pair of the same number underneath the radical, you can take one out of the radical. Therefore, the numerator simplifies to: . Then, get rid of the negative exponent on the denominator (by placing it in the numerator, you get rid of the negative exponent!): $\frac{8x^4y^6z^5(x^6)}{20y^4z^2}$ . Now simplify like terms so that you get: $\frac{2x^{10}y^2z^3}{5}$ .

← Didn't Know|Knew It →

Question

Simplify the following equation:

$\frac{\sqrt{14}}{\sqrt{6}}$

Tap to reveal answer

Answer

You are not allowed to have radicals in the denominator, so to simplify this you must multiply everything by $\sqrt{6}$

So:

$\frac{\sqrt{14}}{\sqrt{6}}\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{84}}{6}$

However, this is not the last step because you need to simplify to numerator, looking for perfect squares.

So:

$\frac{\sqrt{4}\sqrt{21}}{6}=\frac{{2}\sqrt{21}}{6}$

Then you need to simplify a little further to:

$\frac{\sqrt{4}\sqrt{21}}{6}=\frac{{}\sqrt{21}}{3}$

← Didn't Know|Knew It →

Question

Which of the following is equivalent to the expression $\frac{2}{\sqrt{2}}$ ?

Tap to reveal answer

Answer

Which of the following is equivalent to the expression $\frac{2}{\sqrt{2}}$ ?

When dealing with radicals in denominators, we want to bring them up to the numerator. To do so, multiply the beginning fraction by a fraction equal to 1 which looks like the beginning radical (look at the red part of the following expression).

$\frac{2}{\sqrt{2}}{\color{Red} \frac{\sqrt{2}}{\sqrt{2}}}=\frac{2\sqrt{2}}{\sqrt{2}\sqrt{2}}$

Now, instead of canceling the square roots, multiply them out and then simplify.

$\frac{2\sqrt{2}}{\sqrt{2}*\sqrt{2}}=\frac{2\sqrt2}{2}=\sqrt{2}$

← Didn't Know|Knew It →

Question

Simplify: $\frac{2}{4+\sqrt{8}}$

Tap to reveal answer

Answer

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

Rationalize the denominator:

$\frac{2}{4+\sqrt{8}}*\frac{4-\sqrt{8}}{4-\sqrt{8}}$

Simplify:

$\frac{2(4-\sqrt{8})}{16-8}$

$\frac{2(4-\sqrt{8})}{8}$

$\frac{(4-\sqrt{8})}{4}$

$\mathbf{1-\frac{\sqrt{8}}{4}}$

← Didn't Know|Knew It →

Question

Rationalize the numerator: $\frac{\sqrt{6}-\sqrt{8}}{2}$

Tap to reveal answer

Answer

To rationalize the numerator you must multiply by the "conjugate" of the numerator. To find the conjugate simply switch the sign of the expression.

$\frac{\sqrt{6}-\sqrt{8}}{2}$

$\frac{\sqrt{6}-\sqrt{8}}{2}\frac{\sqrt{6}+\sqrt{8}}{\sqrt{6}+\sqrt{8}}$

foil the numerator. The middle terms will cancel each other and the radical cancel each other.

$\frac{6-8}{2(\sqrt{6}+\sqrt{8})}$

$\mathbf{\frac{-1}{(\sqrt{6}+\sqrt{8})}}$

← Didn't Know|Knew It →

Question

Simplify the following:

$\small \frac{\sqrt{7}}{\sqrt14}$

Tap to reveal answer

Answer

To rationalize the denominator you need to multiply the top and bottom by the square root of 14. Upon doing so, you get:

$\small \frac{\sqrt{98}}{14}$

However you can further simplify this because you know that two times forty-nine is ninety-eight you can simplify this to:

$\small \frac{7\sqrt{2}}{14}=\frac{1\sqrt2}{2}$

← Didn't Know|Knew It →

Question

Simplify the following radical:

$\sqrt{\frac{x^5}{4x^7}}$

Tap to reveal answer

Answer

To simplify this radical, first we can simplify what is underneath the radical:

$\sqrt{\frac{x^5}{4x^7}}=\sqrt{\frac{x^5}{4\cdot x^5\cdot x^2}}=\sqrt{\frac{1}{4x^2}}$

Finally, we take the square root of top and bottom:

$\frac{\sqrt{1}}{\sqrt{4x^2}}=\frac{1}{2x}$

The square root of 1 is 1, and the square root of $\sqrt{4x^2}=\sqrt{4}\cdot \sqrt{x^2}=2x$ .

← Didn't Know|Knew It →

Question

Simplify:

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

Tap to reveal answer

Answer

In order to simplify this problem we need to multiply the top and bottom terms of the fraction by the radical in the denominator. This is because we always want an integer in the denominator.

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

Simplify.

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

← Didn't Know|Knew It →

Question

Simplify:

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

Tap to reveal answer

Answer

First, we can simplify the expression by dividing the integers on the top and bottom of the fraction.

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

In order to simplify this problem we need to multiply the top and bottom terms of the fraction by the radical in the denominator. This is because we always want an integer in the denominator.

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

Simplify.

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

← Didn't Know|Knew It →

Question

Simplify:

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

Tap to reveal answer

Answer

In order to get rid of the radical on the bottom of the fraction, we need to multiply the top and bottom of the fraction by the conjugate. The conjugate is the expression that possesses the opposite sign of the radical expression on the bottom of the fraction:

$\sqrt{2}+1\rightarrow\sqrt{2}-1$

Therefore:

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

Simplify.

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

← Didn't Know|Knew It →

Question

Simplify:

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

Tap to reveal answer

Answer

In order to get rid of the radical on the bottom of the fraction, we need to multiply the top and bottom of the fraction by the conjugate. The conjugate is the expression that possesses the opposite sign of the radical expression on the bottom of the fraction:

$\sqrt{3}-1\rightarrow \sqrt{3}+1$

Therefore:

$\frac{3}{\sqrt{3}-1}*\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3\sqrt{3}+3}{3-1}$

Simplify.

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

← Didn't Know|Knew It →

0

Knew It