## College Algebra (10th Edition)

$5i$
$\bf{\text{Solution Outline:}}$ 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}