Intermediate Algebra (12th Edition)

$-\dfrac{4k\sqrt{3z}}{z}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $-\sqrt{\dfrac{48k^2}{z}} ,$ multiply the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{48k^2}{z}\cdot\dfrac{z}{z}} \\\\= -\sqrt{\dfrac{48k^2z}{z^2}} .\end{array} Rewriting the radicand using an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{16k^2}{z^2}\cdot3z} \\\\= -\sqrt{\left(\dfrac{4k}{z}\right)^2\cdot3z} .\end{array} Extracting the root of the radicand that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\dfrac{4k}{z}\sqrt{3z} \\\\= -\dfrac{4k\sqrt{3z}}{z} .\end{array}