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$n$ ^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 $b$

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:

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