### Numbers: Powers and roots

### Higher roots

The *root* we looked at earlier is also known as a square root. This enables us to better distinguish it with the higher roots, which we will define here.

Since #\blue2^\green3=\orange8#, we call #\blue2# the #\green{\text{cube}}# root of #\orange8#. We write this as:

\[\sqrt[\green3]{\orange8}=\blue2\]

Similarly, because #\blue2^\green4=\orange{16}#, we call #\blue2# the #\green{\text{fourth}}# root of #\orange{16}#. We write this as:

\[\sqrt[\green4]{\orange{16}}=\blue2\]

In general we can state:

*The #\green{\textit{cube}}# root of a #\orange{\textit{number}}# is a #\blue{\textit{number}}# which, when raised to the power of #\green{3}#, equals #\orange{\textit{the number inside the radical symbol}}#.*

*The #\green{\textit{fourth}}# root of a #\orange{\textit{number}}# is a #\blue{\textit{non-negative number}}# which, when raised to the power of #\green{4}#, equals #\orange{\textit{the number inside the radical symbol}}#.*

We can extend this to even higher roots, noting that if the #\green{\text{index}}# is odd, the #\blue{\text{result}}# can be both positive and negative, and if the #\green{\text{index}}# is even, the #\blue{\text{result}}# can only be non-negative.

**Examples**

\[\begin{array}{rcl}\sqrt[\green3]{\orange{27}}&=&\blue3 \\ &&\text{because }\blue3^\green{3}=\orange{27} \\ \\\sqrt[\green4]{\orange{625}}&=&\blue5 \\&&\text{because }\blue5^\green4=\orange{625} \text{ and } \blue5 \geq 0 \\ \\\sqrt[\green5]{\orange{32}}&=&\blue2 \\&&\text{because }\blue2^\green5=\orange{32}\\ \\\sqrt[\green3]{\orange{-8}}&=&\blue{-2} \\&&\text{because }(\blue{-2})^\green3=\orange{-8} \\ \\\sqrt[\green6]{\orange{729}}&=&\blue3 \\&&\text{because }\blue3^\green6=\orange{729} \text{ and } \blue3 \geq 0 \end{array}\]

Similar to square roots, higher roots often do not result in integers.

According to the definition of the cube root, it holds that: \[\left(\sqrt[\green3]{\orange2}\right)^3=\orange2\]

We would like to estimate the value of #\sqrt[\green3]{\orange2}#.

Because #1^3=1# and #2^3=8#, we see that #1 \lt \sqrt[\green3]{\orange2} \lt 2#.

Using a calculator, we find the approximation #\sqrt[\green3]{\orange2} \approx 1.25992105....#

Similar to #\sqrt{2}#, #\sqrt[\green3]{\orange2}# has an infinite number of decimal places. When the answer to a question should be exact, #\sqrt[\green3]{\orange2}# is considered a final answer. Like #1#, #\tfrac{1}{2}# #0.6# and #\sqrt{2}#, #\sqrt[\green3]{\orange2}# is a number.

We already saw in the examples in the definition that the cube root of a negative number exists. For higher roots with an even power, this does not hold.

According to the definition of the #\green{\text{fourth}}# root, #\sqrt[\green4]{\orange{-1}}# should be a number that, when raised to the power of #\green4#, equals #\orange{-1}#.

However, when we raise a positive number to power of #\green4#, the result is always a positive number. Therefore, a positive number raised to the power of #\green4# can never be #\orange{-1}#.

A negative number raised to the power of #\green4# also always results in a positive number. Therefore, a negative number raised to the power of #\green4# can never result in #\orange{-1}#.

This means that #\sqrt[\green4]{\orange{-1}}# does not exist. We can only take #\green{\text{fourth}}# roots of non-negative numbers. The same applies to all higher roots with an even #\green{\text{index}}#. For higher roots with an odd #\green{\text{index}}#, this does not hold.

**Other examples**

\[\begin{array}{rcl}\\ \sqrt[\green3]{\orange{-64}}&=&\blue{-4} \\ \\ \sqrt[\green4]{\orange{-16}} && \text{does not exist} \\ \\ \sqrt[\green5]{\orange{-243}}&=&\blue{-3} \\ \\ \sqrt[\green6]{\orange{-729}} && \text{does not exist}\end{array}\]

When we calculate #\sqrt[6]{15625}#, we are looking for a non-negative number which, when raised to the power of #6#, equals #15625#.

In this case:

\[5^6=15625\]

Therefore, #\sqrt[6]{15625}=5#

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