Example14.3.2
Simplify \(\sqrt{y^2+2y+1}\text{.}\)
Solution
\begin{align*}
\sqrt{y^2+2y+1}\amp=\sqrt{(y+1)^2}\\
\amp=\abs{y+1}
\end{align*}
We used the property \(\sqrt[n]{x^n}=\abs{x}\text{ where }n\text{ is even}\text{.}\)
Section14.3Radical Expression Operations
¶In <<Unresolved xref, reference "section-simplifying-square-root"; check spelling or use "provisional" attribute>> and <<Unresolved xref, reference "section-square-root-operations"; check spelling or use "provisional" attribute>>, we learned square root operations. In this section, we will extend those concepts into root of other degrees, and learn how to handle radical expressions with variables.
Subsection14.3.1Simplifying Radical Expressions
In <<Unresolved xref, reference "section-simplifying-square-root"; check spelling or use "provisional" attribute>>, we learned how to convert expressions like \(\sqrt{12}=2\sqrt{3}\text{.}\) In this subsection, we will use the same concept to simplify expressions like \(\sqrt[3]{16x^3}\text{.}\)
We learned this property for square root:
\begin{equation*} \sqrt{x^2}=x\quad\text{where }x\ge0 \end{equation*}The condition \(x\ge0\) must be true because this property doesn't hold if \(x\lt0\text{.}\) For example: \(\sqrt{(-1)^2}=\sqrt{1}=1\text{,}\) where \(\sqrt{(-1)^2}\ne-1\text{.}\) We will use a different way to write this property:
\begin{equation*} \sqrt{x^2}=\abs{x} \end{equation*}This new formula will take care of negative numbers, like in \(\sqrt{(-1)^2}=\abs{-1}\text{.}\)
We have learned
\begin{equation*} \sqrt[n]{x}=x^{\frac{1}{n}} \end{equation*}and it's easy to see
\begin{align*} \sqrt{x^2}\amp=(x^2)^{\frac{1}{2}}\\ \amp=x^{2\cdot\frac{1}{2}}\\ \amp=\abs{x} \end{align*}Similarly, we can show the following:
\begin{align*} \sqrt{x^2}\amp=\abs{x}\\ \sqrt[4]{x^4}\amp=\abs{x}\\ \sqrt[6]{x^6}\amp=\abs{x}\\ \quad\ldots\\ \sqrt[n]{x^n}\amp=\abs{x}\amp\text{where }n\text{ is even} \end{align*}However, for \(\sqrt[3]{x^3}=x\text{,}\) there is no need for the absolute value symbol, because \(\sqrt[3]{1^3}=1\) and \(\sqrt[3]{(-1)^3}=-1\) do not cause confusions. The same is true for other odd-number-degree roots. In summary, we have:
\begin{align*} \sqrt[m]{x^m}\amp=x\amp\text{where }m\text{ is odd}\\ \sqrt[n]{x^n}\amp=\abs{x}\amp\text{where }n\text{ is even} \end{align*}Let's look at a few examples on simplifying radicals.
Example14.3.3
Simplify \(\sqrt{z^4}\text{.}\)
Solution
\begin{align*}
\sqrt{z^4}\amp=\sqrt{z^2z^2}\\
\amp=\abs{z}\cdot\abs{z}\\
\amp=z^2
\end{align*}
In the last example, there is no need to write \(\sqrt{z^4}=\abs{z^2}\text{.}\) Because \(z^2\) is always positive or zero, the absolute value symbols are redundant.
In general, we have
\begin{equation*} \sqrt{x^{2m}}=\abs{x^m}\quad\text{for example: }\sqrt{x^{12}}=x^6 \end{equation*}Note that \(\abs{x^6}\) is not necessary, because \(x^6\) is never negative. Similarly, we have:
\begin{align*} \sqrt[3]{x^{3m}}\amp=x^m\amp\text{for example: }\sqrt[3]{x^{12}}=x^4\\ \sqrt[4]{x^{4m}}\amp=\abs{x^m}\amp\text{for example: }\sqrt[4]{x^{12}}=\abs{x^3}\\ \ldots \end{align*}Example14.3.4
Simplify \(\sqrt{72m^3n^5}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\sqrt{72m^3n^5}\amp=\sqrt{2^33^2m^3n^5}\\
\amp=\sqrt{2\cdot2^23^2m\cdot m^2\cdot n\cdot n^4}\\
\amp=2\cdot3\cdot mn^2\cdot\sqrt{2mn}\\
\amp=6mn^2\cdot\sqrt{2mn}
\end{align*}
Since all variables are assumed positive, there is no need to use the absolute value symbols.
Example14.3.5
Simplify \(\sqrt[3]{16z^4}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\sqrt[3]{16z^4}\amp=\sqrt[3]{2^4z^4}\\
\amp=\sqrt[3]{2^3\cdot2\cdot z^3\cdot z}\\
\amp=2z\cdot\sqrt[3]{2z}
\end{align*}
We used the property \(\sqrt[m]{x^m}=x\text{ where }m\text{ is odd}\text{.}\)
Remark14.3.6
Note that the "dot" (multiplication symbol) in \(2z\cdot\sqrt[3]{2z}\) is necessary. Otherwise it's difficult to tell between \(2z\cdot\sqrt[3]{2z}\) and \(2z^3\cdot\sqrt{2z}\text{.}\)
Example14.3.7
Simplify \(\sqrt[3]{54x^3y^5}\text{.}\)
Solution
\begin{align*}
\sqrt[3]{54x^3y^5}\amp=\sqrt[3]{2\cdot3^3x^3y^3y^2}\\
\amp=3xy\cdot\sqrt[3]{2y^2}
\end{align*}
Subsection14.3.2Exercises
Simplifying Radicals
Subsection14.3.3Radical Operations
In <<Unresolved xref, reference "section-square-root-operations"; check spelling or use "provisional" attribute>>, we learned how to add/subtract/multiply square root expressions. We will extend those concepts and simplify more complicated radical expressions.
In <<Unresolved xref, reference "theorem-square-root-property"; check spelling or use "provisional" attribute>>, we learned the properties:
\begin{align*} \sqrt{x\cdot y}\amp=\sqrt{x}\cdot\sqrt{y}\\ \sqrt{\frac{x}{y}}\amp=\frac{\sqrt{x}}{\sqrt{y}}, \text{ where }y\ne0 \end{align*}Those properties also work for other radicals:
\begin{align*} \sqrt[n]{x\cdot y}\amp=\sqrt[n]{x}\cdot\sqrt[n]{y}\\ \sqrt[n]{\frac{x}{y}}\amp=\frac{\sqrt[n]{x}}{\sqrt[n]{y}}, \text{ where }y\ne0 \end{align*}The next few examples will use those properties.
Example14.3.9
Do multiplication: \(\sqrt[3]{4x^2y}\cdot2\cdot\sqrt[3]{4xy^2}\)
Solution
\begin{align*}
\sqrt[3]{4x^2y}\cdot2\cdot\sqrt[3]{4xy^2}\amp=2\cdot\sqrt[3]{4x^2y\cdot4xy^2}\\
\amp=2\cdot\sqrt[3]{2^2x^2y\cdot2^2xy^2}\\
\amp=2\cdot\sqrt[3]{2^3\cdot2x^3y^3}\\
\amp=2\cdot2xy\cdot\sqrt[3]{2}\\
\amp=4xy\cdot\sqrt[3]{2}
\end{align*}
Example14.3.10
Do multiplication: \(\sqrt{2p^2-p-1}\cdot\sqrt{p^2-1}\)
Solution
\begin{align*}
\sqrt{2p^2-p-1}\cdot\sqrt{p^2-1}\amp=\sqrt{(2p+1)(p-1)}\cdot\sqrt{(p+1)(p-1)}\\
\amp=\sqrt{(2p+1)(p-1)(p+1)(p-1)}\\
\amp=\abs{p-1}\sqrt{(2p+1)(p+1)}
\end{align*}
Example14.3.11
Do division: \(\frac{\sqrt[4]{32m^6n^{10}}}{\sqrt[4]{162m^{10}n}}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\frac{\sqrt[4]{32m^6n^{10}}}{\sqrt[4]{162m^{10}n}}\amp=\sqrt[4]{\frac{32m^6n^{10}}{162m^{10}n}}\\
\amp=\sqrt[4]{\frac{16n^9}{81m^4}}\\
\amp=\sqrt[4]{\frac{2^4n^9}{3^4m^4}}\\
\amp=\frac{2n^2\cdot\sqrt[4]{n}}{3m}
\end{align*}
When we add/subtract radical expressions, remember we can only combine like terms.
Example14.3.12
Simplify \(\sqrt{8t}+\sqrt[3]{8t}+\sqrt{18t}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\sqrt{8t}+\sqrt[3]{16t}+\sqrt{18t}\amp=\sqrt{2^3t}+\sqrt[3]{2^4t}+\sqrt{2\cdot3^2t}\\
\amp=2\sqrt{2t}+2\cdot\sqrt[3]{2t}+3\sqrt{2t}\\
\amp=5\sqrt{2t}+2\cdot\sqrt[3]{2t}
\end{align*}
Note that \(\sqrt{2t}\) and \(\sqrt[3]{2t}\) are not like terms.
Example14.3.13
Simplify \(\sqrt[3]{375n^{12}x^9}-\sqrt[3]{192n^{12}x^9}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\sqrt[3]{375n^{12}x^9}-\sqrt[3]{192n^{12}x^9}\amp=\sqrt[3]{3\cdot5^3n^{12}x^9}-\sqrt[3]{2^6 3n^{12}x^9}\\
\amp=5n^4x^3\cdot\sqrt[3]{3}-2^2n^4x^3\cdot\sqrt[3]{3}\\
\amp=5n^4x^3\cdot\sqrt[3]{3}-4n^4x^3\cdot\sqrt[3]{3}\\
\amp=n^4x^3\cdot\sqrt[3]{3}
\end{align*}
When we add/subtract fractions, don't forget to find common denominators first.
Example14.3.14
Simplify \(\sqrt{\frac{3}{4}}+\sqrt{12}\text{.}\) Assume all variables are positive.
Solution
\begin{align*}
\sqrt{\frac{3}{4}}+\sqrt{12}\amp=\frac{\sqrt{3}}{\sqrt{4}}+\sqrt{2^2\cdot3}\\
\amp=\frac{\sqrt{3}}{2}+2\sqrt{3}\\
\amp=\frac{\sqrt{3}}{2}+\frac{2\sqrt{3}}{1}\\
\amp=\frac{\sqrt{3}}{2}+\frac{\multiplyleft{2}2\sqrt{3}}{\multiplyleft{2}1}\\
\amp=\frac{\sqrt{3}}{2}+\frac{4\sqrt{3}}{2}\\
\amp=\frac{\sqrt{3}+4\sqrt{3}}{2}\\
\amp=\frac{5\sqrt{3}}{2}
\end{align*}
It's important to review radical operations covered in Subsection 8.2.7.
Subsection14.3.4Exercises
Multiplying/Dividing Radicals
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Mixed Operations of Radicals