Nth Root

Similarly to square root and cube root, we can define nth root.

Nth root of a number $b$ is such number $a$, that $a^n=b$.

Notation for the nth root is the following: $\color{purple}{\sqrt[n]{b}}$.

Square root is the 2nd root, which is $\sqrt{b}$ (we just don't write 2).

Cube root is 3rd root, which is ${\sqrt[{{3}}]{{{b}}}}$.

Fourth root is ${\sqrt[{{4}}]{{{b}}}}$.

And so forth.

Nth root symbol $\color{purple}{\sqrt[n]{\phantom{0}}}$ is a radical with a small $n$.

This is done to emphasize the fact, that we are looking for a number, that when raised to nth power) will give original number.

Let's go through a couple of examples.

Example 1. Find ${\sqrt[{{4}}]{{{2401}}}}$.

It is known that ${{7}}^{{4}}={2401}$, so ${\sqrt[{{4}}]{{{2401}}}}={7}$.

If ${n}$ is odd, then we can find nth root of negative number as well.

Example 2. Find ${\sqrt[{{5}}]{{-{32}}}}$.

Since ${{\left(-{2}\right)}}^{{5}}=-{32}$, then ${\sqrt[{{5}}]{{-{32}}}}=-{2}$.

Is it possible, to find nth root of a number if ${n}$ is even?

Example 3. Find ${\sqrt[{{8}}]{{-{256}}}}$.

What is ${\sqrt[{{8}}]{{-{256}}}}$. It is such number ${a}$, that ${{a}}^{{8}}=-{256}$.

But can we really find such number, that when raised to even power, will give negative number?

Answer is no, because any number, raised to even power will give positive number.

For, example ${{\left(-{3}\right)}}^{{4}}={\left(-{3}\right)}\cdot{\left(-{3}\right)}\cdot{\left(-{3}\right)}\cdot{\left(-{3}\right)}={81}$ (there is even number of minuses, so they just disappear).

As a result, we have the following useful fact.

Nth root of negative number (if ${n}$ is even) doesn't exist.

Finally, notice that nth root undoes raising to nth power (if we take nth root of a number and then raise the result to nth power, we will get back to the original number), and vice versa (with a slight modification):

${\color{red}{{{{\left({\sqrt[{{n}}]{{{b}}}}\right)}}^{{n}}={b}}}}$

${\color{green}{{{\sqrt[{{n}}]{{{{b}}^{{n}}}}}={b}}}}$, if ${n}$ is odd

${\color{purple}{{{\sqrt[{{n}}]{{{{b}}^{{n}}}}}={\left|{b}\right|}}}}$, if ${n}$ is even

Did you notice absolute value? Why is that?

Because raising any number to even number will give positive number, and taking nth root of the result will also give positive number. Absolute value guarantees, that we wil have positive number.

We could write, that ${{\left({\sqrt[{{n}}]{{{b}}}}\right)}}^{{n}}={\left|{b}\right|}$, but that is not necessary, because if ${n}$ is even, ${b}$ is already non-negative number (n-th root of negative number doesn't exist, if ${n}$ is even).

Let's go through a couple of example.

Example 1. ${{\left({\sqrt[{{4}}]{{{15}}}}\right)}}^{{4}}={15}$.

Example 2. ${{\left({\sqrt[{{3}}]{{-{2}}}}\right)}}^{{3}}=-{2}$.

Example 3. ${{\left({\sqrt[{{8}}]{{-{5}}}}\right)}}^{{8}}$ doesn't exist.

Example 4. ${\sqrt[{{3}}]{{{{\left({\color{red}{{-{0.2}}}}\right)}}^{{3}}}}}={\sqrt[{{3}}]{{-{0.008}}}}={\color{red}{{-{0.2}}}}$.

Example 5. ${\sqrt[{{10}}]{{{{\left({\color{red}{{-{2}}}}\right)}}^{{10}}}}}={\sqrt[{{10}}]{{{1024}}}}={2}={\left|{\color{red}{{-{2}}}}\right|}$.