Question

express in simplest radical form. \\(\sqrt{108}\\)

Explanation:

Step1: Factor 108 into prime factors

First, we factorize 108. We know that \(108 = 36\times3\), and \(36 = 6\times6 = 2\times3\times2\times3=2^{2}\times3^{2}\). So, \(108 = 2^{2}\times3^{2}\times3\).

Step2: Apply the square - root property

The square - root property states that \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for \(a\geq0,b\geq0\)) and \(\sqrt{a^{2}} = a\) (for \(a\geq0\)).
We have \(\sqrt{108}=\sqrt{2^{2}\times3^{2}\times3}\).
Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\), we can rewrite it as \(\sqrt{2^{2}}\times\sqrt{3^{2}}\times\sqrt{3}\).
Since \(\sqrt{2^{2}} = 2\) and \(\sqrt{3^{2}}=3\), then \(\sqrt{2^{2}}\times\sqrt{3^{2}}\times\sqrt{3}=2\times3\times\sqrt{3}\).

Step3: Simplify the expression

Simplify \(2\times3\times\sqrt{3}\), we get \(6\sqrt{3}\).

Answer:

\(6\sqrt{3}\)