Question

question 29 of 40
simplify the radical. assume that all variables represent positive re
\sqrt{49p^{3}q^{4}}
\sqrt{49p^{3}q^{4}} =

Explanation:

Step1: Factor into perfect squares

We can rewrite the radicand \(49p^{3}q^{4}\) as \(49\times p^{2}\times p\times q^{4}\), where \(49 = 7^{2}\), \(p^{2}\) is a perfect square, and \(q^{4}=(q^{2})^{2}\) is a perfect square. So, \(\sqrt{49p^{3}q^{4}}=\sqrt{7^{2}\times p^{2}\times p\times(q^{2})^{2}}\).

Step2: Apply square - root property

Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for \(a\geq0,b\geq0\)) and \(\sqrt{a^{2}} = a\) (for \(a\geq0\)), we have:
\(\sqrt{7^{2}\times p^{2}\times p\times(q^{2})^{2}}=\sqrt{7^{2}}\times\sqrt{p^{2}}\times\sqrt{q^{2\times2}}\times\sqrt{p}\)
Since \(\sqrt{7^{2}} = 7\), \(\sqrt{p^{2}}=p\) (because \(p>0\)) and \(\sqrt{(q^{2})^{2}}=q^{2}\), we get \(7\times p\times q^{2}\times\sqrt{p}\)
Simplifying the expression, we have \(7pq^{2}\sqrt{p}\)

Answer:

\(7pq^{2}\sqrt{p}\)