Answer
$5i$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\sqrt{(4+3i)(3i-4)}
,$ 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{(3i+4)(3i-4)}
\\\\=
\sqrt{(3i)^2-(4)^2}
\\\\=
\sqrt{9i^2-16}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{9(-1)-16}
\\\\=
\sqrt{-9-16}
\\\\=
\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}
