Quartic and higher roots in standard form

Calculating with numbers: Calculating with powers and roots

Quartic and higher roots in standard form

We will discuss higher roots, but we start with a special case, namely the quartic root. Like a square, you can only take a quartic root of a nonnegative number.

The quartic root

The quartic root or fourth root of a number, $a\ge 0$ is by definition the nonnegative integer $w$ such that $w^4 = a$ . Notation: $w = \sqrt[4]{a}$ and $w=a^{\frac{1}{4}}$ .

Examples

$\begin{aligned}\sqrt[4]{16}&={2}\quad\text{because }2^4=16\text{ and }\\[0.2cm] \sqrt{81}&={3}\quad\text{because }(3)^4=81\end{aligned}$

The rules for calculating quartic roots resemble those of square roots.

Quartic root of a fourth power

For any natural number $n$ we have: $\sqrt[4]{n^4}=n$ and $\left(\sqrt[4]{n}\right)^4=n$

Example $n=3$

$\sqrt[4]{3^4}=3$ and $\left(\sqrt[4]{3}\right)^4=3$

Product rule of roots

For natural number $m$ and $n$ applies: $\sqrt[4]{m}\times \sqrt[4]{n}= \sqrt[4]{m\times n}$

example

$\sqrt[4]{2}\times \sqrt[4]{8}=\sqrt[4]{2\times 8}$ because $\begin{aligned}\left(\sqrt[4]{2}\times \sqrt[4]{8}\right)^4 &= \left(\sqrt[4]{2}\right)^4\times \left(\sqrt[4]{8}\right)^4\\ &= 2\times 8\\ &= 16\\ &= \left(\sqrt[4]{16}\right)^4 \\ &= \left(\sqrt[4]{2\times 8}\right)^4\end{aligned}$

Sum rule? How temping it may be, there is no sum rule for quartic roots. For nonnegative numbers $m$ and $n$ we have: $\sqrt[4]{m}+\sqrt[4]{n}\;\red{\neq}\;\sqrt[4]{m+n}$ A concrete example with numbers illustrates this: $5=3+2=\sqrt[4]{81}+\sqrt[4]{16}\;\red{\neq}\;\sqrt[4]{81+16}=\sqrt[4]{97}$

Quotient rule of quartic roots

The quartic root of a fraction with positive natural numbers in the numerator and the denominator is equal to the quotient of the quartic root of the numerator and the quartic root of the denominator .

In formula language we have for positive natural numbers $m$ and $n$ : $\sqrt[4]{\frac{m}{n}}=\frac{\sqrt[4]{m}}{\sqrt[4]{n}}$

example

$\sqrt[4]{\frac{81}{216}}=\frac{\sqrt[4]{81}}{\sqrt[4]{216}}=\frac{3}{4}$ and indeed $\left(\frac{3}{4}\right)^4=\frac{3^4}{4^4}=\frac{81}{216}$

But there is a new rule for a quartic root regarding the reduction of quartic root to square root under special circumstances.

Reduction rule of quartic roots

For any natural number $n$ we have: $\sqrt[4]{n^2}= \sqrt{n}$

Example

$\sqrt[4]{9}=\sqrt[4]{3^2}=\sqrt{3}$ because $\begin{aligned}\left(\sqrt[4]{9}\right)^4 &= 9\\ &= 3^2\\ &= \bigl((\sqrt{3})^2\bigr)^2\\ &= \sqrt{3}^4 \end{aligned}$

The above rules can be used to simplify quartic roots.

The quartic root of a natural number greater than 1, say $\sqrt[4]{n}$ , is called irreducible if $n$ has no quartic number (i.e., a fourth power of a natural number) greater than 1 as divisor and also cannot be reduced to a square root because the number under root sign is a square number. So $\sqrt[4]{24}=\sqrt[3]{2^3\times 3}$ is an irreducible quartic root, but $\sqrt{36}$ and $\sqrt[4]{1250}$ are not irreducible, because $\begin{aligned}\sqrt[4]{36}&=\sqrt[4]{6^2}\\ &=\sqrt{6}\\ \\ \sqrt[4]{1250} &=\sqrt[4]{625\times 2}\\ &=\sqrt[4]{5^4\times 2}\\ &=\sqrt[4]{5^4}\times\sqrt[4]{2}\\ &=5\times \sqrt[4]{2}\end{aligned}$ The last expression we usually write shorter as $5\sqrt[4]{2}$ .

Every quartic root of a natural number greater than 1 can be written in standard form, i.e., as a natural number, a square root, or as the product of a natural number, and an irreducible square or quartic root.

The expression $m\sqrt[4]{n}$ for integers $m$ and $n>1$ is in standard form if

there exists no fourth power of a natural number greater than 1 that divides $n$ , and
$n$ is not equal to a square number.

You find the standard form of a quartic root by 'extracting all fourth powers from the quartic root' and by reducing the quartic root of a square number to a square root. The following examples illustrate this.

Write $\sqrt[4]{736}$ in standard form.

$\sqrt[4]{736}={}$ $2\sqrt[4]{46}$

First we find the largest possible fourth power that divides $736$ .
In this case we can write: $\begin{aligned}736&=16\times 46\\ \\ &={2}^4\times 46\end{aligned}$ This example follows from the prime factorisation of $736$ : $736=2^5\times 23$ Once the largest fourth power has been found that divides $736$ we apply the calculation rules $\sqrt[4]{m\times n} = \sqrt[4]{m}\times\sqrt[4]{n}$ and $\sqrt[4]{n^4}=n$ for natural numbers $m$ and $n$ to: $\begin{aligned} \sqrt[4]{736} &= \sqrt[4]{{2}^4\times 46}\\ \\ &=\sqrt[4]{{2}^4}\times \sqrt[4]{46}\\ \\ &=2\sqrt[4]{46}\end{aligned}$

New example

Now that we know how square, cube and quartic roots can be treated mathematically, the way for higher power roots lies open.

In general, the $n$ -th root $\sqrt[n]{a}$ of $a$ is the number $w$ such that $w^n =a$ , provided that $a\ge 0$ in case $n$ is even. If $n$ is even, then we have $w^n=(-w)^n$ and there are two candidates for the root: the conventional choice is the positive number. We denote this root also as $a^{\frac{1}{n}}$ .

$\begin{aligned} \sqrt[n]{a}\times \sqrt[n]{b}&=\sqrt[n]{a\times b}\\ \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}&=\sqrt[n]{\frac{a}{b}}\\ \\ \left(\sqrt[n]{a}\right)^m &= \sqrt[n]{a^m}\\ \\ \left(\sqrt[n]{a}\right)^n &=a \\ \\ \sqrt[m\times n]{a^{m}} &= \sqrt[n]{a}\end{aligned}$ If $n$ is even, these rules only hold for positive values of $a$ and $b$ .

Standard form of higher power roots

The expression $\frac{a}{b}\sqrt[n]{c}$ , where $a$ , $b$ and $c$ are positive natural numbers, is called a standard form of a higher root of a positive rational number if

$\frac{a}{b}$ is an irreducible fraction,
$c$ has no $n$ -th power other than $1$ is divisor,
$c$ is not equal to a $d$ -th power of each divisor $d$ of $n$

The third condition is new compared to the cases $n=2$ and $n=3$ and is connected with the last of the above calculation rules.

For example, the standard form of $\sqrt[6]{144}$ is $\sqrt[3]{12}$ because $144 = 2^4\times 3^2=(2^2\times 3)^2 = 18^2$ so $\sqrt[6]{144}=\sqrt[6]{12^2}=\sqrt[3]{12}$ .

Write $\sqrt[3]{77760}$ in standard form.

You can use the following prime factorisation: $77760={2}^{6}\times {3}^{5}\times {5}$

$\sqrt[3]{77760}={}$ $12 \sqrt[3]{45}$

$\begin{aligned} \sqrt[3]{77760} &=\sqrt[3]{{2}^{6}\times {3}^{5}\times {5}} & \blue{\text{prime factorisation}}\\ \\ &= {2}^{2}\times 3\times \sqrt[3]{ {} {{3}^{2}\times } 5 } & \blue{\text{third power extracted from the root}} \\ \\ &= 12 \sqrt[3]{45} & \blue{\text{standard form}}\end{aligned}$

New example