Algebra II : Radicals and Fractions

Study concepts, example questions & explanations for Algebra II

varsity tutors app store varsity tutors android store

Example Questions

Example Question #171 : Simplifying Radicals

Simplify the following expression:

\(\displaystyle \frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}\)

Possible Answers:

\(\displaystyle \frac{1}{4}\)

 

\(\displaystyle \sqrt{4}\)

\(\displaystyle -\frac{1}{4}\)

\(\displaystyle \frac{1}{\sqrt{4}}\)

Correct answer:

\(\displaystyle \frac{1}{4}\)

 

Explanation:

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

\(\displaystyle \frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\)

and

\(\displaystyle \sqrt{a}\sqrt{b}=\sqrt{ab}\)

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

\(\displaystyle \frac{\sqrt{27}}{2}\cdot \sqrt{\frac{1}{108}}\)

\(\displaystyle \frac{\sqrt{27}}{\sqrt{4}}\cdot \sqrt{\frac{1}{108}}\)

\(\displaystyle \sqrt{\frac{27}{4}}\cdot \sqrt{\frac{1}{108}}\)

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

\(\displaystyle \sqrt{\frac{27}{4}\cdot\frac{1}{108}}=\sqrt{\frac{27}{4\cdot 108}}\)

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:

\(\displaystyle \sqrt{\frac{3\cdot \mathbf{9}}{4\cdot \mathbf{9}\cdot 12}}=\sqrt{\frac{3}{4\cdot 12}}\)

We also know that 3 is a factor of 12:

\(\displaystyle \sqrt{\frac{\mathbf{3}}{4\cdot 4\cdot \mathbf{3}}}=\sqrt{\frac{1}{4\cdot 4}}\)

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

\(\displaystyle \sqrt{a\cdot a}=\sqrt{a^{2}}=a\)

We can finally simplify this expression completely:

\(\displaystyle \sqrt{\frac{1}{4\cdot 4}}=\frac{\sqrt{1}}{\sqrt{4\cdot 4}}=\frac{1}{4}\)

Example Question #1 : Radicals And Fractions

Rationalize the denominator.

\(\displaystyle \frac{x^{2}}{\sqrt{11}}\)

Possible Answers:

\(\displaystyle \frac{x^{2}}{11}\)

\(\displaystyle 11x^{4}\)

\(\displaystyle \frac{x^{4}\sqrt{11}}{11}\)

\(\displaystyle \frac{x^{2}\sqrt{11}}{11}\)

\(\displaystyle \frac{x^{2}\sqrt{11}}{\sqrt{11}}\)

Correct answer:

\(\displaystyle \frac{x^{2}\sqrt{11}}{11}\)

Explanation:

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

To do this, we multiply the radical by \(\displaystyle \frac{\sqrt{11}}{\sqrt{11}}\),

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

Example Question #2 : Radicals And Fractions

Simplify by rationalizing the denominator:

\(\displaystyle \frac{14}{8 - \sqrt{15}}\)

Possible Answers:

None of the other responses is correct.

\(\displaystyle \frac{16+2 \sqrt{15}}{7}\)

\(\displaystyle \frac{16-14 \sqrt{15}}{7}\)

\(\displaystyle \frac{16-2 \sqrt{15}}{7}\)

\(\displaystyle \frac{16+14 \sqrt{15}}{7}\)

Correct answer:

\(\displaystyle \frac{16+2 \sqrt{15}}{7}\)

Explanation:

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

\(\displaystyle \frac{14}{8 - \sqrt{15}}\)

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

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

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

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

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

\(\displaystyle = \frac{16+2 \sqrt{15}}{7}\)

Example Question #261 : Radicals

Simplify: 

\(\displaystyle \sqrt{\frac{9}{16}}\)

Possible Answers:

\(\displaystyle \frac{3}{4}\)

\(\displaystyle \frac{3}{8}\)

\(\displaystyle \frac{\sqrt{9}}{8}\)

\(\displaystyle \frac{3}{16}\)

Correct answer:

\(\displaystyle \frac{3}{4}\)

Explanation:

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

\(\displaystyle \sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}\)

Example Question #5 : Radicals And Fractions

Simplify the following:

\(\displaystyle \frac{4\sqrt{5}}{\sqrt{20}}\)

Possible Answers:

\(\displaystyle \pm2\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \sqrt{5}\)

\(\displaystyle \frac{4\sqrt{5}}{5}\)

\(\displaystyle \frac{\sqrt{4}}{5}\)

Correct answer:

\(\displaystyle \pm2\)

Explanation:

You can begin by rewriting this equation as:

\(\displaystyle 4 * \frac{\sqrt{5}}{\sqrt{20}}=4 * \sqrt{\frac{5}{20}} = 4 * \sqrt{\frac{1}{4}}=\frac{4}{\sqrt{4}}\)

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

\(\displaystyle \frac{4}{\sqrt{4}} * \frac{\sqrt{4}}{\sqrt{4}} = \frac{4\sqrt{4}}{4}\)

Then, cancel the common \(\displaystyle 4\):

\(\displaystyle \frac{4\sqrt{4}}{4} = \sqrt{4}\)

Since \(\displaystyle 4\) is a perfect square you can take the square root to get the simplified answer.

\(\displaystyle \sqrt4=\pm 2\) 

Example Question #261 : Radicals

Simplify \(\displaystyle \frac{\sqrt 32}{\sqrt2}\).

Possible Answers:

\(\displaystyle \pm 4\)

None of the other answers.

\(\displaystyle \frac{4}{\sqrt2}\)

\(\displaystyle 16\)

\(\displaystyle 4\sqrt2\)

Correct answer:

\(\displaystyle \pm 4\)

Explanation:

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: \(\displaystyle \sqrt{x}\cdot \sqrt{y} = \sqrt {xy}\)

 

For this equation:

 

\(\displaystyle \sqrt{\frac{32}{2}} = \sqrt{16} = \pm4\)

Example Question #7 : Radicals And Fractions

Solve and simplify.

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

Possible Answers:

\(\displaystyle 9+\sqrt{5}\)

\(\displaystyle 6\sqrt{5}\)

\(\displaystyle 6+\sqrt{5}\)

\(\displaystyle 14+6\sqrt{5}\)

\(\displaystyle 14\)

Correct answer:

\(\displaystyle 14+6\sqrt{5}\)

Explanation:

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.

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

\(\displaystyle =9+3\sqrt{5}+3\sqrt{5}+5\)

Combining like terms we get our final answer as follows.

\(\displaystyle =14+6\sqrt{5}\)

Example Question #8 : Radicals And Fractions

Solve and simplify.

\(\displaystyle \frac{9}{4\sqrt{5}}\)

Possible Answers:

\(\displaystyle \frac{9}{20}\)

\(\displaystyle \frac{4\sqrt{5}}{5}\)

\(\displaystyle 45\)

\(\displaystyle \frac{9\sqrt{5}}{20}\)

\(\displaystyle \frac{3\sqrt{5}}{10}\)

Correct answer:

\(\displaystyle \frac{9\sqrt{5}}{20}\)

Explanation:

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 \(\displaystyle \sqrt{5}\). 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 \(\displaystyle 5\).

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

Final answer is \(\displaystyle \frac{9\sqrt{5}}{20}\).

Example Question #9 : Radicals And Fractions

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

Possible Answers:

\(\displaystyle x^{10}y^2z^3}\)

\(\displaystyle \frac{2x^{10}y^2z^3}{5}\)

\(\displaystyle \frac{2xy^2z^3}{5}\)

\(\displaystyle \frac{x^{10}y^2z^3}{5}\)

\(\displaystyle \frac{2x^{10}y^2z^3}{3}\)

Correct answer:

\(\displaystyle \frac{2x^{10}y^2z^3}{5}\)

Explanation:

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: \(\displaystyle 8x^4y^6z^5\). Then, get rid of the negative exponent on the denominator (by placing it in the numerator, you get rid of the negative exponent!): \(\displaystyle \frac{8x^4y^6z^5(x^6)}{20y^4z^2}\). Now simplify like terms so that you get: \(\displaystyle \frac{2x^{10}y^2z^3}{5}\).

Example Question #10 : Radicals And Fractions

Simplify \(\displaystyle \sqrt{116x^{11}y^4z^7}\)

Possible Answers:

\(\displaystyle 2x^5y^2z\sqrt{29xz}\)

\(\displaystyle 2x^3y^2z^3\sqrt{29xz}\)

\(\displaystyle 2x^5y^2z^3\sqrt{29xz}\)

\(\displaystyle 2x^5y^2z^3\sqrt{xz}\)

\(\displaystyle 4x^5y^2z^3\sqrt{29xz}\)

Correct answer:

\(\displaystyle 2x^5y^2z^3\sqrt{29xz}\)

Explanation:

To simplify radicals, I like to approach each term separately. I would start by doing a factor tree for \(\displaystyle 116\), so you can see if there are any pairs of numbers that you can take out. \(\displaystyle 116\)factors to \(\displaystyle 29\cdot 2\cdot 2\), so you can take a \(\displaystyle 2\) out of the radical. For \(\displaystyle x^{11}\), there are \(\displaystyle 5\) pairs of \(\displaystyle x\)'s, so \(\displaystyle x^{5}\) goes outside of the radical, and one \(\displaystyle x\) remains underneath the radical. For \(\displaystyle y^4\), there are \(\displaystyle 2\) pairs of \(\displaystyle y\)'s, so you can take \(\displaystyle 2\) \(\displaystyle y\)'s outside the radical. For \(\displaystyle z^7\), there are \(\displaystyle 3\) complete pairs of \(\displaystyle z\)'s so \(\displaystyle z^3\) goes on the outside, while one \(\displaystyle z\) remains underneath the radical. Now, put those all together to get: \(\displaystyle 2x^5y^2z^3\sqrt{29xz}\).

Learning Tools by Varsity Tutors