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Section 4.6 Radical Expressions
Subsection 4.6.1 Reducing Radical Expressions
Radical expressions are algebraic expressions involving
\(n\) -th roots. See
Section 3.2 .) Often we want to reduce radical expressions involving powers. This can be done by recognizing powers of
\(n\) under the radical first and then extracting the base outside the radical. For example,
\(\sqrt[3]{32} = \sqrt[3]{2^3\cdot 2^2} = 2\sqrt[3]{2^2} = 2 \sqrt[3]{4}\text{.}\) Care must be taken when dealing with even versus odd
\(n\) -th roots. In particular,
\(\sqrt[n]{x^n} = \left|x\right|\text{,}\) for
even positive integers
\(n\text{.}\) See
Principle 3.2.6 .
Example 4.6.1 . Reducing Radical Expressions (Even).
Reduce the radical expression, keeping in mind that the \(x\) could be negative.
\begin{equation*}
\sqrt[4]{32 x^6} = \sqrt[4]{2^4 \cdot 2 \cdot x^4 \cdot x^2} = 2 |x| \sqrt[4]{2 x^2}.
\end{equation*}
You can see the need for absolute value by considering a negative example of say \(x=-1\text{.}\) In this case,
\begin{equation*}
\sqrt[4]{32 (-1)^6} = \sqrt[4]{32} = 2 \sqrt[4]{2}
\end{equation*}
must be positive. Without out the absolute value, it would not be.
Example 4.6.2 . Reducing Radical Expressions (Odd).
Reduce the radical expression, keeping in mind that the \(x\) could be negative.
\begin{equation*}
\sqrt[3]{32 x^4} = \sqrt[3]{2^3 \cdot 2^2 \cdot x^3 \cdot x} = 2 x \sqrt[3]{4x}.
\end{equation*}
Example 4.6.3 . Reducing Radical Expressions.
Reduce each radical expression, being considerate of the domain.
\(\displaystyle \displaystyle \sqrt{72 x^{9} y^6}\)
\(\displaystyle \displaystyle \sqrt[3]{72 x^{9} y^6}\)
\(\displaystyle \displaystyle \sqrt[4]{72 x^{9} y^6}\)
Solution .
\(\displaystyle \sqrt{72 x^9 y^6}\text{.}\) Observe that \(x\geq 0\text{,}\) for otherwise the expression under the radical would be negative and that’s undefined for a square root. On the other hand, \(y\) may be any real number since \(y^6 \geq 0\) is always the case.
\begin{align*}
\amp= \sqrt{8\cdot 9 \cdot x^{9} y^6}\\
\amp= \sqrt{2^2 \cdot 2 \cdot 3^2 \cdot \left(x^4\right)^2 \cdot x \cdot \left(y^3\right)^2}\\
\amp= 2 \cdot 3 \cdot x^4 \left|y\right|^3\sqrt{ 2 x}\\
\amp= 6 \cdot x^4 \left|y\right|^3\sqrt{ 2 x}
\end{align*}
\(\displaystyle \sqrt[3]{72 x^9 y^6}\text{.}\) Since this is a cube root, the variables may take any real number value.
\begin{align*}
\amp= \sqrt[3]{2^3 \cdot 3^2 \left(x^3\right)^3 \left(y^2\right)^3}\\
\amp= 2x^3 y^2 \sqrt[3]{3^2}\\
\amp= 2x^3 y^2 \sqrt[3]{9}
\end{align*}
\(\displaystyle \sqrt[4]{72 x^9 y^6}\text{.}\) As in the first example, we require \(x \geq 0\) while \(y\) may be any real number.
\begin{align*}
\amp= \sqrt[4]{2^3 \cdot 3^2 \cdot x \cdot x^4 \cdot x^4 \cdot y^4 \cdot y^2 } \\
\amp= |x| |x| |y| \sqrt[4]{2^3 \cdot 3^2 \cdot x y^2 } \\
\amp= x^2 y \sqrt[4]{72 x y^2 }
\end{align*}
We can check our result by raising it to the fourth power and comparing it to the expression under the radical in
\(\displaystyle \sqrt[4]{72 x^9 y^6}\text{:}\)
\begin{equation*}
\left(x^2 y \sqrt[4]{72 x y^2 }\right)^4 = x^8 y^4 \cdot 72 x y^2 = 72 x^9 y^6.
\end{equation*}
Subsection 4.6.2 Rationalizing Radical Expressions
Another common technique is to multiply by a form of one to rearrange radical expressions. For instance, it’s common to remove square roots from the denominator (or numerator) by multiplying by a form of one
\begin{equation*}
\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times {\color{blue}\frac{\sqrt{2}}{\sqrt{2}}} = \sqrt{2}{(\sqrt{2})^2} = \frac{\sqrt{2}}{2}.
\end{equation*}
This process is called rationalizing the denominator. We can use something similar when working with algebraic expressions involving radicals.
Example 4.6.4 . Addition with Radicals.
Consider the sum
\begin{equation*}
\frac{\sqrt{1-x^2}}{x}+\frac{2}{\sqrt{1-x^2}}
\end{equation*}
Find a common denominator to make the addition possible.
\begin{align*}
\frac{\sqrt{1-x^2}}{x}+&\frac{2}{\sqrt{1-x^2}}\\
&= \frac{\sqrt{1-x^2}}{x} \cdot {\color{blue}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}} + \frac{2}{\sqrt{1-x^2}} \cdot {\color{blue}\frac{x}{x}} \\
&= \frac{(\sqrt{1-x^2})^2+2x}{x\sqrt{1-x^2}}\\
&= \frac{1-x^2+2x}{x\sqrt{1-x^2}}\\
&= \frac{-x^2+2x+1}{x\sqrt{1-x^2}}
\end{align*}
Note that you may not cancel the factor of \(x\) in the denominator as it is not a factor of the numerator.
Example 4.6.5 . Multiplication with Radicals.
The expression below looks complicated.
\begin{equation*}
\frac{\sqrt{2x-y}-2}{2x-y-4}\cdot \frac{\sqrt{2x-y}+2}{\sqrt{2x-y}+2}
\end{equation*}
But multiplication results in a difference of squares (see
(4.4.2) ) in the numerator. Observe the difference in sign and the cancellation of the middle terms.
\begin{align*}
\frac{\sqrt{2x-y}-2}{2x-y-4}\cdot & \frac{\sqrt{2x-y}+2}{\sqrt{2x-y}+2}\\
&= \frac{\left(\sqrt{2x-y}\right)^2+{\color{blue} 2\sqrt{2x-y}}-{\color{blue}2\sqrt{2x-y}}-2^2}{\left(2x-y-4\right)\left(\sqrt{2x-y}+2\right)}\\
&= \frac{{\color{green} 2x-y-4}}{\left({\color{green}2x-y-4}\right)\left(\sqrt{2x-y}+2\right)}\\
&= \frac{1}{\sqrt{2x-y}+2}
\end{align*}
In the last example have multiplied the left factor by a form of one which leaves it unchanged.
\begin{equation*}
\frac{\sqrt{2x-y}-2}{2x-y-4} = \frac{\sqrt{2x-y}-2}{2x-y-4} \cdot \frac{\sqrt{2x-y}+2}{\sqrt{2x-y}+2} = \frac{1}{\sqrt{2x-y}+2}
\end{equation*}
This is a useful technique used in calculus also referred to rationalizing .
Example 4.6.6 . Rationalizing the Numerator.
Rationalize the numerator and simplify
\begin{equation*}
\frac{\sqrt{2+h}-\sqrt{2}}{h}.
\end{equation*}
Solution .
\begin{align*}
\frac{\sqrt{2+h}-\sqrt{2}}{h} &\\
&= \frac{\sqrt{2+h}-\sqrt{2}}{h} \cdot \frac{\sqrt{2+h}+\sqrt{2}}{\sqrt{2+h}+\sqrt{2}}\\
&= \frac{2+h-2}{h\left(\sqrt{2+h}+\sqrt{2}\right)}\\
&= \frac{h}{h\left(\sqrt{2+h}+\sqrt{2}\right)}\\
&= \frac{1}{\sqrt{2+h}+\sqrt{2}}
\end{align*}
