#
$n$
^{
th
}
Roots

A square root of a number $b$ , written $\sqrt{b}$ , is a solution of the equation ${x}^{2}=b$ .

Example: $\sqrt{49}=7$ , because ${7}^{2}=49$ .

Similarly, the
**
cube root
**
of a number
$b$
, written
$\sqrt[3]{b}$
, is a solution to the equation
${x}^{3}=b$
.

Example: $\sqrt[3]{64}=4$ , because ${4}^{3}=64$ .

More generally, the ${n}^{\text{th}}$ root of $b$ , written $\sqrt[n]{b}$ , is a number $x$ which satisfies ${x}^{n}=b$ .

The ${n}^{\text{th}}$ root can also be written as a fractional exponent:

$\sqrt[n]{b}={b}^{\frac{1}{n}}$

## When does the ${n}^{\text{th}}$ root exist, and how many are there?

If you are working in the real number system only, then

- If $n$ is an even whole number, the ${n}^{\text{th}}$ root of $b$ exists whenever $b$ is positive ; and for all $b$ .
- If $n$ is an odd whole number, the ${n}^{\text{th}}$ root of $b$ exists for all $b$

**
Examples:
**

$\sqrt[4]{-81}$ is not a real number.

$\sqrt[5]{-32}=-2$

If you are working in the complex number system , then things get more, well, complex.

Here
*
every
*
number has
$2$
square roots,
$3$
cube roots,
$4$
fourth roots,
$5$
fifth roots, etc.

For example, the $4$ fourth roots of the number $81$ are $3,-3,3i$ and $-3i$ . Because:

$\begin{array}{l}{3}^{4}=81\\ {\left(-3\right)}^{4}=81\\ {\left(3i\right)}^{4}={3}^{4}{i}^{4}=81\\ {\left(-3i\right)}^{4}={\left(-3\right)}^{4}{i}^{4}=81\end{array}$