Difficulty Level: At Grade Created by: CK-12

Radicals are the roots of values. In fact, the word radical comes from the Latin word “radix,” meaning “root.” You are most comfortable with the square root symbol \begin{align*}\sqrt{x}\end{align*}; however, there are many more radical symbols.

A radical is a mathematical expression involving a root by means of a radical sign.

\begin{align*}\sqrt[3]{y}=x && \text{because} \ x^3=y && \sqrt[3]{27}=3, \ because \ 3^3=27\\ \sqrt[4]{y}=x && \text{because} \ x^4=y && \sqrt[4]{16}=2 \ because \ 2^4=16\\ \sqrt[n]{y}=x && \text{because} \ x^n=y && \end{align*}

Some roots do not have real values; in this case, they are called undefined.

Even roots of negative numbers are undefined.

\begin{align*}\sqrt[n]{x}\end{align*} is undefined when \begin{align*}n\end{align*} is an even whole number and \begin{align*}x<0\end{align*}.

Example 1: Evaluate the following radicals:

• \begin{align*}\sqrt[3]{64}\end{align*}
• \begin{align*}\sqrt[4]{-81}\end{align*}

Solution: \begin{align*}\sqrt[3]{64} = 4\end{align*} because \begin{align*}4^3=64\end{align*}

\begin{align*}\sqrt[4]{-81}\end{align*} is undefined because \begin{align*}n\end{align*} is an even whole number and \begin{align*}-81<0\end{align*}.

In Chapter 8, you learned how to evaluate rational exponents:

\begin{align*}a^{\frac{x}{y}} \ where \ x=power \ and \ y=root\end{align*}

This can be written in radical notation using the following property.

Rational Exponent Property: For integer values of \begin{align*}x\end{align*} and whole values of \begin{align*}y\end{align*}:

\begin{align*}a^{\frac{x}{y}}= \sqrt[y]{a^x}\end{align*}

Example: Rewrite \begin{align*}x^{\frac{5}{6}}\end{align*} using radical notation.

Solution: This is correctly read as the sixth root of \begin{align*}x\end{align*} to the fifth power. Writing in radical notation, \begin{align*}x^{\frac{5}{6}}=\sqrt[6]{x^5}\end{align*}, where \begin{align*}x^5>0\end{align*}.

Example 2: Evaluate \begin{align*}\sqrt[4]{4^2}\end{align*}.

Solution: This is read, “The fourth root of four to the second power.”

\begin{align*}4^2=16\end{align*}

The fourth root of 16 is 2; therefore,

\begin{align*}\sqrt[4]{4^2}=2\end{align*}

In Chapter 1, Lesson 5, you learned how to simplify a square root. You can also simplify other radicals, like cube roots and fourth roots.

Example: Simplify \begin{align*}\sqrt[3]{135}\end{align*}.

Solution: Begin by finding the prime factorization of 135. This is easily done by using a factor tree.

\begin{align*}&\sqrt[3]{135}= \sqrt[3]{3 \cdot 3 \cdot 3 \cdot 5} = \sqrt[3]{3^3} \cdot \sqrt[3]{5}\\ & 3 \sqrt[3]{5}\end{align*}

\begin{align*}a \sqrt[n]{x}+b\sqrt[n]{x}=(a+b)\sqrt[n]{x}\end{align*}

Example 3: Add \begin{align*}3\sqrt{5}+6\sqrt{5}\end{align*}.

Solution: The value “\begin{align*}\sqrt{5}\end{align*}” is considered a like term. Using the rule above,

\begin{align*}3 \sqrt{5}+6\sqrt{5}=(3+6) \sqrt{5}=9\sqrt{5}\end{align*}

Example: Simplify \begin{align*}2\sqrt[3]{13} + 6 \sqrt[3]{12}\end{align*}.

Solution: The cube roots are not like terms, therefore there can be no further simplification.

In some cases, the radical may need to be reduced before addition/subtraction is possible.

Example 4: Simplify \begin{align*}4\sqrt{3}+2\sqrt{12}\end{align*}.

Solution: \begin{align*}\sqrt{12}\end{align*} simplifies as \begin{align*}2\sqrt{3}\end{align*}.

\begin{align*}4\sqrt{3}+2\sqrt{12} &\rightarrow 4\sqrt{3}+2\left ( 2\sqrt{3} \right )\\ 4\sqrt{3}+4\sqrt{3}&=8\sqrt{3}\end{align*}

To multiply radicands, the roots must be the same.

\begin{align*}\sqrt[n]{a} \cdot \sqrt[n]{b}= \sqrt[n]{ab}\end{align*}

Example: Simplify \begin{align*}\sqrt{3} \cdot \sqrt{12}\end{align*}.

Solution: \begin{align*}\sqrt{3} \cdot \sqrt{12}=\sqrt{36}=6\end{align*}

Dividing radicals is more complicated. A radical in the denominator of a fraction is not considered simplified by mathematicians. In order to simplify the fraction, you must rationalize the denominator.

To rationalize the denominator means to remove any radical signs from the denominator of the fraction using multiplication.

Remember: \begin{align*}\sqrt{a} \times \sqrt{a}= \sqrt{a^2}=a\end{align*}

Example 1: Simplify \begin{align*}\frac{2}{\sqrt{3}}\end{align*}.

Solution: We must clear the denominator of its radical using the property above. Remember, what you do to one piece of a fraction, you must do to all pieces of the fraction.

\begin{align*}\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3^2}}=\frac{2\sqrt{3}}{3}\end{align*}

Example: Simplify \begin{align*}\frac{7}{\sqrt[3]{5}}\end{align*}.

Solution: In this case, we need to make the number inside the cube root a perfect cube. We need to multiply the numerator and the denominator by \begin{align*}\sqrt[3]{5^2}\end{align*}.

\begin{align*}\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}=\frac{7^3\sqrt{25}}{\sqrt[3]{5^3}}=\frac{7^3\sqrt{25}}{5}\end{align*}

Example: A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square-feet. Find the dimensions of the pool and the area of the pool.

Solution:

1. Make a sketch.
2. Let \begin{align*}x=\end{align*} the width of the pool.
3. Write an equation. \begin{align*}Area=length \cdot width\end{align*}

Combined length of pool and walkway \begin{align*}=2x+2\end{align*}

Combined width of pool and walkway \begin{align*}=x+2\end{align*}

\begin{align*}\text{Area}=(2x+2)(x+2)\end{align*}

Since the combined area of pool and walkway is \begin{align*}400 \ ft^2\end{align*} we can write the equation.

\begin{align*}(2x+2)(x+2)=400\end{align*}

4. Solve the equation:

\begin{align*}&& & (2x+2)(x+2)=400\\ & \text{Multiply in order to eliminate the parentheses}. && 2x^2+4x+2x+4=400\\ & \text{Collect like terms}. && 2x^2+6x+4=400\\ & \text{Move all terms to one side of the equation}. && 2x^2+6x-396=0\\ & \text{Divide all terms by} \ 2. && x^2+3x-198=0\end{align*}

\begin{align*}x & = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & = \frac{-3 \pm \sqrt{3^2-4(1)(-198)}}{2(1)}\\ & = \frac{-3\pm \sqrt{801}}{2} \approx \frac{-3\pm 28.3}{2}\end{align*}

Use the Quadratic Formula. \begin{align*}x \approx 12.65\end{align*} or –15.65 feet

5. We can disregard the negative solution since it does not make sense for this context. Thus, we can check our answer of 12.65 by substituting the result in the area formula.

\begin{align*}\text{Area} = [2(12.65)+2)](12.65+2)=27.3 \cdot 14.65 \approx 400 \ ft^2.\end{align*}

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. For which values of \begin{align*}n\end{align*} is \begin{align*}\sqrt[n]{-16}\end{align*} undefined?

1. \begin{align*}\sqrt{169}\end{align*}
2. \begin{align*}\sqrt[4]{81}\end{align*}
3. \begin{align*}\sqrt[3]{-125}\end{align*}
4. \begin{align*}\sqrt[5]{1024}\end{align*}

Write each expression as a rational exponent.

1. \begin{align*}\sqrt[3]{14}\end{align*}
2. \begin{align*}\sqrt[4]{zw}\end{align*}
3. \begin{align*}\sqrt{a}\end{align*}
4. \begin{align*}\sqrt[9]{y^3}\end{align*}

Write the following expressions in simplest radical form.

1. \begin{align*}\sqrt{24}\end{align*}
2. \begin{align*}\sqrt{300}\end{align*}
3. \begin{align*}\sqrt[5]{96}\end{align*}
4. \begin{align*}\sqrt{\frac{240}{567}}\end{align*}
5. \begin{align*}\sqrt[3]{500}\end{align*}
6. \begin{align*}\sqrt[6]{64x^8}\end{align*}
7. \begin{align*}\sqrt[3]{48a^3b^7}\end{align*}
8. \begin{align*}\sqrt[3]{\frac{16x^5}{135y^4}}\end{align*}
9. True or false? \begin{align*}\sqrt[7]{5} \cdot \sqrt[6]{6}=\sqrt[42]{30}\end{align*}

Simplify the following expressions as much as possible.

1. \begin{align*}3\sqrt{8}-6\sqrt{32}\end{align*}
2. \begin{align*}\sqrt{180}+6\sqrt{405}\end{align*}
3. \begin{align*}\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}\end{align*}
4. \begin{align*}\sqrt{8x^3}-4x\sqrt{98x}\end{align*}
5. \begin{align*}\sqrt{48a}+\sqrt{27a}\end{align*}
6. \begin{align*}\sqrt[3]{4x^3}+x\sqrt[3]{256}\end{align*}

Multiply the following expressions.

1. \begin{align*}\sqrt{6}\left ( \sqrt{10} + \sqrt{8} \right )\end{align*}
2. \begin{align*}\left ( \sqrt{a} - \sqrt{b} \right ) \left ( \sqrt{a} + \sqrt{b} \right )\end{align*}
3. \begin{align*}\left ( 2\sqrt{x}+ 5 \right ) \left ( 2\sqrt{x}+5 \right )\end{align*}

Rationalize the denominator.

1. \begin{align*}\frac{7}{\sqrt{15}}\end{align*}
2. \begin{align*}\frac{9}{\sqrt{10}}\end{align*}
3. \begin{align*}\frac{2x}{\sqrt{5}x}\end{align*}
4. \begin{align*}\frac{\sqrt{5}}{\sqrt{3}y}\end{align*}
5. The volume of a spherical balloon is \begin{align*}950 cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}=\frac{4}{3} \pi R^3\end{align*})
6. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 in^2\end{align*}, what is the width of the frame?
7. The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.

Mixed Review

1. An item originally priced \begin{align*}\c\end{align*} is marked down 15%. The new price is 612.99. What is \begin{align*}c\end{align*}? 2. Solve \begin{align*}\frac{x+3}{6}=\frac{21}{x}\end{align*}. 3. According to the Economic Policy Institute (EPI), minimum wage in 1989 was3.35 per hour. In 2009, it was 7.25 per hour. What is the average rate of change? 4. What is the vertex of \begin{align*}y=2(x+1)^2+4\end{align*}? Is this a minimum or a maximum? 5. Using the minimum wage data (adjusted for inflation) compiled from EPI, answer the following questions. 1. Graph the data as a scatter plot. 2. Which is the best model for this data: linear, quadratic, or exponential? 3. Find the model of best fit and use it to predict minimum wage adjusted for inflation for 1999. 4. According to EPI, the 1999 minimum wage adjusted for inflation was6.58. How close was your model?
5. Use interpolation to find minimum wage in 1962.
Year Minimum Wage Adj. for Inflation Year Minimum Wage Adj. for Inflation
1947 3.40 1952 5.36
1957 6.74 1960 6.40
1965 7.52 1970 7.81
1978 7.93 1981 7.52
1986 6.21 1990 6.00
1993 6.16 1997 6.81
2000 6.37 2004 5.80
2006 5.44 2008 6.48
2009 7.25

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