Question

simplify.
\sqrt{4v^{21}}
assume that the variable represents a positive real number.
your answer

Explanation:

Step1: Recall square - root property

We know that for a non - negative real number \(a\) and \(b\), \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)). We want to simplify \(\sqrt{4v^{21}}\). First, we can split the radicand (the expression inside the square root) into factors that are perfect squares.
We know that \(4 = 2^{2}\) and \(v^{21}=v^{20}\cdot v=(v^{10})^{2}\cdot v\) (using the property of exponents \(a^{m + n}=a^{m}\cdot a^{n}\) and \((a^{m})^{n}=a^{mn}\)).
So, \(\sqrt{4v^{21}}=\sqrt{4\times v^{20}\times v}\)

Step2: Apply the square - root property

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), we have:
\(\sqrt{4\times v^{20}\times v}=\sqrt{4}\times\sqrt{v^{20}}\times\sqrt{v}\)
Since \(\sqrt{4} = 2\) (because \(2^{2}=4\)) and \(\sqrt{v^{20}}=v^{10}\) (because \((v^{10})^{2}=v^{20}\) and \(v\) is a positive real number), we substitute these values in:
\(\sqrt{4}\times\sqrt{v^{20}}\times\sqrt{v}=2\times v^{10}\times\sqrt{v}\)
Simplifying the product, we get \(2v^{10}\sqrt{v}\)

Answer:

\(2v^{10}\sqrt{v}\)