Answer
$5i$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\sqrt{(3+4i)(4i-3)}
,$ use special products to simplify the radicand. Take note that $i^2=-1.$
$\bf{\text{Solution Details:}}$
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{(4i+3)(4i-3)}
\\\\=
\sqrt{(4i)^2-(3)^2}
\\\\=
\sqrt{16i^2-9}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{16(-1)-9}
\\\\=
\sqrt{-16-9}
\\\\=
\sqrt{-25}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{-1}\cdot\sqrt{25}
.\end{array}
Since $i=\sqrt{-1},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
i\sqrt{25}
.\end{array}
Extracting the root of the factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
i\sqrt{(5)^2}
\\\\=
i(5)
\\\\=
5i
.\end{array}
