# $n$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:

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