# Back to our Roots

I am tired of fractions, you too? Well let’s switch gears and talk about the roots of numbers. This is preparing you for a post or posts on the rules of exponents.

So before, I introduced the concept of the square root and how it is the reverse operation of squaring a number:

$\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm{5}$ because ${5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}$. That is $\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}$. Or in general, $\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}$. Remember that when taking the square root, there are two solutions since ${\left({{-}{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}$ as well.

Well, what about the opposite operation to ${x}^{3}$? Well there is one:

$\sqrt{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}$

Notice a few things here. First, there is only one solution. There are no plus/minus solutions because the index (the “3”) of the root is odd and using the rules of signs, the sign of the odd root of a number will be the same as the number in the radical symbol. That is:

$\sqrt{125}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}\sqrt{{-}{125}}\hspace{0.33em}{=}\hspace{0.33em}{-}{5}$

The other thing to notice is that if there is no index shown, then a “2” is assumed to be there. If the index is a “3”, it is called a cube root. After that, you use ordinal numbers, that is fourth root, fifth root, etc.

The last thing to notice is if the index is even, then you do get two plus/minus solutions. If the index is odd, you only get one solution.

So in general,

$\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}$ if n is even and

$\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}$ if n is odd.

Examples:

$\begin{array}{l} {\sqrt{16}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}\pm{2}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{2}}\\ {\sqrt{{-}{8}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}{-}{2}{)}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{-}{8}}\\ {\sqrt{32}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{2}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\ {\sqrt{{-}{32}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}{-}{2}{)}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{-}{2}}\\ {\sqrt{81}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}\pm{3}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{3}} \end{array}$

In my next post, I’ll introduce some rules regarding exponents.

Posted on Categories Algebra, Pre-VCE