Standard notation of higher roots

Numbers: Powers and roots

Standard notation of higher roots

As with square roots, we can write higher roots in standard notation using the calculation rules.

When there is a higher root in an answer, we simplify it to the standard notation. This means that the higher root needs to meet the following requirements:

We bring the greatest possible factor in front of the radical sign.
This means that we remove any numbers to the power of the index of the root, from the number inside the radical sign.
\[\sqrt[\green3]{16}=\sqrt[\green3]{2^\green3 \times 2}=\sqrt[\green3]{2^\green3} \times \sqrt[\green3]{2}=2\sqrt[\green3]{2}\]
We make sure there are no higher roots in the denominator of the fraction.
We do this by multiplying the numerator and the denominator of the fraction by a number to make sure the denominator does not contain a higher root anymore. A smart choice for this number is the higher power from the denominator, exponentiated to the index of the root minus #1#.
\[\frac{1}{\sqrt[\green4]{3}}=\frac{\left(\sqrt[\green4]{3}\right)^{\green4-1}}{\sqrt[\green4]{3} \times \left(\sqrt[\green4]{3}\right)^{\green4-1}}=\frac{\left(\sqrt[\green4]{3}\right)^3}{3}\]
We make sure the index of the higher root is as small as possible.
We do this by checking if the number inside the radical sign can be written as a power with an exponent that is a divisor of the index of the root.
\[\sqrt[\green8]{324}=\sqrt[\green8]{18^2}=\sqrt[4]{18}\]

Examples

\[\begin{array}{rcl}\sqrt[\green4]{96}&=&\sqrt[\green4]{2^\green4 \times 6} \\ &=& \sqrt[\green4]{2^\green4} \times \sqrt[\green4]{6} \\ &=& 2 \sqrt[\green4]{6} \\ \\\sqrt[\green5]{15552}&=&\sqrt[\green5]{2^\green5 \times 3^\green5 \times 2} \\&=& \sqrt[\green5]{2^\green5} \times \sqrt[\green5]{3^\green5} \times \sqrt[\green5]{2} \\ &=& 2 \times 3 \times \sqrt[\green5]{2} \\ &=& 6 \sqrt[\green5]{2} \\ \\ \dfrac{2}{\sqrt[\green3]{5}}&=& \dfrac{2 \left(\sqrt[\green3]{5}\right)^2}{\sqrt[\green3]{5} \times \left(\sqrt[\green3]{5}\right)^2}\\ &=& \dfrac{2 \left(\sqrt[\green3]{5}\right)^2}{5} \\ \\ \sqrt[\green6]{225}&=& \sqrt[\green6]{15^2}\\&=&\sqrt[3]{15}\end{array}\]

In practice

In practice we address the first requirement by writing the number in its prime factorization, and checking if there are any prime factors with a power greater than or equal to the index of the root. If this is the case, we can remove any factors with an exponent equal to the index, from the root.

Write \(\sqrt[4]{161462109375}\) in standard notation.

You can use the fact that #161462109375={3}^{10}\times {5}^{8}\times {7}#.

#\sqrt[4]{161462109375}=# #225 \sqrt[4]{63}#

#\begin{array}{rcl}
\sqrt[4]{161462109375}&=& \sqrt[4]{{3}^{10}\times {5}^{8}\times {7}} \\ &&\phantom{xxx}\blue{\text{number inside the radical symbol written in prime factorization}} \\
&=& \sqrt[4]{\left({3}^{2}\right)^{4} \times {3}^{2}\times \left({5}^{2}\right)^{4} \times {7}} \\ &&\phantom{xxx}\blue{\text{powers of }4\text{ isolated using the calculation rules for powers}}\\
&=& \sqrt[4]{\left({3}^{2}\right)^{4} } \times \sqrt[4]{\left({5}^{2}\right)^{4}} \times \sqrt[4]{ {{3}^{2}\times } {} 7 } \\ &&\phantom{xxx}\blue{\text{calculation rule: higher root of a product equals the product of higher roots}} \\
&=& {3}^{2}\times {5}^{2}\times \sqrt[4]{ {{3}^{2}\times } {} 7 } \\ &&\phantom{xxx}\blue{\text{eliminated root(s)}} \\
&=& 225 \sqrt[4]{63}\\ &&\phantom{xxx}\blue{\text{calculated}}
\end{array}#

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