2. Radical notation

Let \(a > 0\) be a real number and '\(n\)' be a positive integer. Then \(\sqrt[n]{a} = b\), if \(b^n = a\) and \(b > 0\).

 

The symbol $\sqrt[n]{}$ is called a radical, \(n\) is the index of the radical and \(a\) is called the radicand.

 

Let us find the root when \(n\) is odd.

 

When \(n = 3\) and \(a = 125\), we have \(b^3 = 125\) \(\Rightarrow b = \sqrt[3]{125}\) \(\Rightarrow b = 5\)

 

Therefore, when \(n\) is odd, there is only one real root.

 

Consider finding the root, when \(n\) is even.

 

When \(n = 2\) and \(a = 81\), we have \(b^2 = 81\) \(\Rightarrow b = \sqrt[2]{81}\)

 

In this case, we arrive at \(2\) conclusions, \(9 \times 9 = 81\) and \((-9) \times (-9) = 81\). And, we can say \(9\) and \(-9\) are roots of \(b\).

 

But it is incorrect to write that \(\sqrt{81} = \pm{9}\) because when \(n\) is even, the roots are positive only if the radical symbol is positive. That is, \(\sqrt[n]{a}\). Similarly, the roots are negative if the radical symbol is denoted by \(- \sqrt[n]{a}\).

 

Therefore, we need to write \(\sqrt[2]{81} = 9\) and \(- \sqrt[2]{81} = -9\).

Consider the radical notation \(\sqrt[n]{a} = b\) and \(\sqrt[3]{125} = 5\)

 

In this example, the root index tell us that the number of times the number(\(5\)) multiplies itself to gives the radicand(\(125\)).

 

Now, we are going to learn one more way of representing the powers and roots which involves the use of fractional indices.

 

In fractional index, we can write \(\sqrt[n]{a} = b\) as \(a = b^{\frac{1}{n}}\)

In the above example (\(3\)), we can see that the result is \(4^{\frac{5}{2}}\). We say that this is of the form \(x^{\frac{m}{n}}\) which is either \(n^{th}\) root of the \(m^{th}\) power of \(x\) or \(m^{th}\) power of the \(n^{th}\) root of \(x\).

 

That is, \(x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}\) or \((x^n)^{\frac{1}{m}} = \sqrt[n]{x^m}\) or \((\sqrt[n]{x})^m\)