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Section 3.6 Radical Expressions
Radical expressions are algebraic expressions involving
\(n\) -th roots. (See
Section 1.8 .) One common technique is to multiply by a form of one to rearrange the expression. 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 3.25 . 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 3.26 . 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
(3.2) ) 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 3.27 . 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*}
