College Algebra (11th Edition)

$\dfrac{k\sqrt{k}+3k}{k-9}$
$\bf{\text{Solution Outline:}}$ Rationalize the given expression, $\dfrac{k}{\sqrt{k}-3} ,$ by multiplying the numerator and the denominator by the conjugate of the denominator. $\bf{\text{Solution Details:}}$ By reversing the operator between the terms of the denominator, $\sqrt{k}-3 ,$ the conjugate is $\sqrt{k}+3 .$ Multiplying both the numerator and the denominator by the conjugate results to \begin{array}{l}\require{cancel} \dfrac{k}{\sqrt{k}-3}\cdot\dfrac{\sqrt{k}+3}{\sqrt{k}+3} \\\\= \dfrac{k(\sqrt{k}+3)}{(\sqrt{k}-3)(\sqrt{k}+3)} .\end{array} 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\begin{array}{l}\require{cancel} \dfrac{k(\sqrt{k}+3)}{(\sqrt{k})^2-(3)^2} \\\\= \dfrac{k(\sqrt{k}+3)}{k-9} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{k\sqrt{k}+3k}{k-9} .\end{array}