Let's start with something "easy":
The square root of $x$ is a number such that when we square it we get $x$.
Example: $\sqrt{9}$ is a positive number such that when we square it we get $9.$
In general, $y=\sqrt{x}$ is (almost) the same as saying $y^2=x.$
A Little Bit Harder
The cube root of $x$ is a number such that when we cube it we get $x$.
Example $\sqrt[3]{-8}$ is a number such that when we cube it we get $-8$.
In general, $y=\sqrt[3]{x}$ really is the same as saying $y^3=x.$
The Big Picture: General Radicals
The $n$th root of $x$ is a number such that when we raise it to the $n$th power we get $x$.
In general, $y=\sqrt[n]{x}$ is (ALMOST) the same as $y^n=x.$
Vocab: $n$ is called the index of the radical.
Vocab: The quantity under the radical $x$ is called the radicand.
Example : $\sqrt[6]{64}$ is a number such that when we raise it to the $6$th power we get $64$.
Here the index is $6$ and the radicand is $64.$
Examples: Evaluate the radical expressions.
$\sqrt{36}$
$\sqrt[3]{27}$
$\sqrt[3]{64}+\sqrt[3]{-1}$
$\sqrt[4]{81}$
$\sqrt[5]{32}$
Weird Examples: Evaluate the radical expressions.
$\sqrt{-9}$
$-\sqrt{9}$
$\sqrt[3]{0}$
$\sqrt[4]{1}$
$\sqrt[5]{-1}$
Simplifying Variable Expressions with Roots
We now begin to consider radical expressions which involve variables.
It is tempting to say that $\sqrt{x^2}=x.$
This is one of those "almost true" kinds of statements.
For $x=2,$ the statement is true since $\sqrt{x^2}=\sqrt{2^2}=\sqrt{4}=2=x.$ $\checkmark$
For $x=-2,$ the statement is false since $\sqrt{x^2}=\sqrt{(-2)^2}=\sqrt{4}=2\neq -2=x.$ $\times$
Simplifying Variable Expressions with Roots
We can make an always-true statement by using absolute values: $$\sqrt{x^2}=|x|.$$ This statement is true for both $x=2$ and $x=-2.$
For $x=2,$ the statement is true since $\sqrt{x^2}=\sqrt{2^2}=\sqrt{4}=2=|2|=|x|.$ $\checkmark$
For $x=-2,$ the statement is true since $\sqrt{x^2}=\sqrt{(-2)^2}=\sqrt{4}=2=|-2|=|x|.$ $\checkmark$
This statement is true for any real number $x!$
Simplifying Variable Expressions with Roots
We now begin to consider radical expressions which involve variables.
It is tempting to say that $\sqrt[3]{x^3}=x.$
And by giving in to this temptation, you would not be committing any kind of mathematical naughtiness since...
For $x=2,$ the statement is true since $\sqrt[3]{x^3}=\sqrt[3]{2^3}=\sqrt[3]{8}=2=x.$ $\checkmark$
For $x=-2,$ the statement is true since $\sqrt[3]{x^3}=\sqrt[3]{(-2)^3}=\sqrt[3]{-8}=-2=x.$ $\checkmark$
The General Pattern for Even Indices
$$\sqrt{x^2}=\sqrt[2]{x^2}=|x|$$ $$\sqrt[4]{x^4}=|x|$$ $$\sqrt[6]{x^6}=|x|$$ $$\sqrt[8]{x^8}=|x|$$ $$\vdots$$
The General Pattern for Odd Indices
$$\sqrt[3]{x^3}=x$$ $$\sqrt[5]{x^5}=x$$ $$\sqrt[7]{x^7}=x$$ $$\sqrt[9]{x^9}=x$$ $$\vdots$$
The General Pattern
If $n$ is an even index, then $$\sqrt[n]{x^n}=|x|,$$ and if $n$ is an odd index, then $$\sqrt[n]{x^n}=x.$$
Note for Homework
If the homework says "assume variables represent positive real numbers," you may assume (without guilt!) that $$\sqrt{x^2}=x$$ $$\sqrt[4]{x^4}=x$$ $$\sqrt[6]{x^6}=x$$ $$\sqrt[8]{x^8}=x$$ $$\vdots$$
Examples: Simplifying Variable Expressions with Roots
Method 1: The Power Method.
$\sqrt{64x^2}$
$$
\begin{array}{lll}
\displaystyle \sqrt{64x^2}&\displaystyle=\sqrt{(8x)^2} &\mbox{}\\
\displaystyle &\displaystyle=|8x| &\mbox{even radical}\\
\displaystyle &\displaystyle=|8||x| &\mbox{}\\
\displaystyle &\displaystyle=8|x| &\mbox{}\\
\end{array}
$$
$\sqrt[3]{8 x^{30}y^{12}}$
$$
\begin{array}{lll}
\displaystyle \sqrt[3]{8 x^{30}y^{12}}&\displaystyle=\sqrt[3]{\left(8 x^{10}y^{4}\right)^3} &\mbox{}\\
\displaystyle &\displaystyle=8 x^{10}y^{4} &\mbox{}\\
\end{array}
$$
$-\sqrt[4]{81p^{28}q^{16}}$
$$
\begin{array}{lll}
\displaystyle -\sqrt[4]{81p^{28}q^{16}}&\displaystyle=-\sqrt[4]{\left(3p^{7}q^{4}\right)^4} &\mbox{}\\
\displaystyle &\displaystyle=-|3p^{7}q^{4}| &\mbox{even radical}\\
\displaystyle &\displaystyle=-|3||p^{7}||q^{4}| &\mbox{}\\
\displaystyle &\displaystyle=-3|p^{7}|q^{4} &\mbox{since $|q^{4}|=q^4$}\\
\end{array}
$$
$\sqrt[5]{243n^{25}}$
$$
\begin{array}{lll}
\displaystyle \sqrt[5]{243n^{25}}&\displaystyle=\sqrt[5]{\left(3n^{5}\right)^5} &\mbox{}\\
\displaystyle &\displaystyle=3n^5 &\mbox{}\\
\end{array}
$$
Summary of the Power Method
Step 1: Write the radicand as some expression written to the power of the index of the radical. You may need to recall some exponent rules to do this correctly.
Step 2: If #1) the index is even AND #2) the variables can be any real number, put your final expression in absolute value signs.
If the directions say "assume variables represent positive real numbers," then you don't need to worry about absolute value signs.
Examples: Simplifying Variable Expressions with Roots
This time, "assume variables represent positive real numbers."
Method 2: The jail-break method. Groups of the size of the index can escape from radical jail. Sadly, only one member of the group can make it out alive.
$-\sqrt{100 a^2 b^6}$
$$
\begin{array}{lll}
\displaystyle -\sqrt{100 a^2 b^6}&\displaystyle=-\sqrt{2 \cdot 2 \cdot 5 \cdot 5 \cdot a \cdot a \cdot b \cdot b \cdot b \cdot b \cdot b \cdot b} &\mbox{factor completely}\\
\displaystyle &\displaystyle=-\sqrt{\underline{2 \cdot 2} \cdot \underline{5 \cdot 5} \cdot \underline{ a \cdot a } \cdot \underline{b \cdot b} \cdot \underline{b \cdot b} \cdot \underline{b \cdot b}} &\mbox{underline pairs}\\
\displaystyle &\displaystyle=-2 \cdot 5 \cdot a \cdot b \cdot b \cdot b &\mbox{each pair sends out a representative}\\
\displaystyle &\displaystyle=-10ab^3 &\mbox{simplify}\\
\end{array}
$$
$\sqrt[3]{125p^{9}q^{6}}$
$$
\begin{array}{lll}
\displaystyle \sqrt[3]{125p^{9}q^{6}} &\displaystyle=\sqrt[3]{5 \cdot 5 \cdot 5 \cdot p \cdot p \cdot p \cdot p \cdot p \cdot p \cdot p \cdot p \cdot p \cdot q \cdot q \cdot q \cdot q \cdot q \cdot q} &\mbox{factor completely}\\
\displaystyle &\displaystyle=\sqrt[3]{\underline{5 \cdot 5 \cdot 5} \cdot \underline{p \cdot p \cdot p} \cdot \underline{p \cdot p \cdot p} \cdot \underline{p \cdot p \cdot p} \cdot \underline{q \cdot q \cdot q} \cdot \underline{q \cdot q \cdot q}} &\mbox{underline triplets}\\
\displaystyle &\displaystyle=5 \cdot p \cdot p \cdot p \cdot q \cdot q &\mbox{each triplet sends out a representative}\\
\displaystyle &\displaystyle=5 p^3 q^2&\mbox{simplify}\\
\end{array}
$$
$\sqrt[4]{81n^{12}}$
$$
\begin{array}{lll}
\displaystyle \sqrt[4]{81n^{12}}&\displaystyle=\sqrt[4]{3 \cdot 3 \cdot 3 \cdot 3 \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n \cdot n } &\mbox{factor completely}\\
\displaystyle &\displaystyle=\sqrt[4]{\underline{3 \cdot 3 \cdot 3 \cdot 3} \cdot \underline{n \cdot n \cdot n \cdot n} \cdot \underline{n \cdot n \cdot n \cdot n} \cdot \underline{n \cdot n \cdot n \cdot n} } &\mbox{underline each quadruplet}\\
\displaystyle &\displaystyle=3 \cdot n \cdot n \cdot n &\mbox{each quadruplet sends out a representative}\\
\displaystyle &\displaystyle=3n^3&\mbox{simplify}\\
\end{array}
$$
$\sqrt[5]{243y^{10}}$
$$
\begin{array}{lll}
\displaystyle \sqrt[5]{243y^{10}} &\displaystyle=\sqrt[5]{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y} &\mbox{factor completely}\\
\displaystyle &\displaystyle=\sqrt[5]{\underline{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3} \cdot \underline{y\cdot y\cdot y\cdot y\cdot y}\cdot \underline{y\cdot y\cdot y\cdot y\cdot y}} &\mbox{underline quintuplets}\\
\displaystyle &\displaystyle=3 \cdot y \cdot y &\mbox{each quintuplet sends out a representative}\\
\displaystyle &\displaystyle=3y^2 &\mbox{simplify}\\
\end{array}
$$
Summary of the Jail-Break Method
Step 1: Factor the radicand completely.
Step 2: Underline groups that are the same size as the index (pairs, triplets, quadruplets, etc.) and "set free" one representative from the radical.
Step 3: If #1) the index is even AND #2) the variables can be any real number, put your final expression in absolute value signs.
If the directions say "assume variables represent positive real numbers," then you don't need to worry about absolute value signs.
Comparing Both Methods
The power method is easier for larger powers, but it is harder to use and remember.
The jail-break method is easier to use and remember, but is very cumbersome for larger powers.