## Intermediate Algebra (12th Edition)

$\dfrac{3x+6\sqrt{xy}}{x-4y}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{3\sqrt{x}}{\sqrt{x}-2\sqrt{y}} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{3\sqrt{x}}{\sqrt{x}-2\sqrt{y}}\cdot\dfrac{\sqrt{x}+2\sqrt{y}}{\sqrt{x}+2\sqrt{y}} \\\\= \dfrac{3\sqrt{x}(\sqrt{x}+2\sqrt{y})}{(\sqrt{x}-2\sqrt{y})(\sqrt{x}+2\sqrt{y})} .\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{3\sqrt{x}(\sqrt{x}+2\sqrt{y})}{(\sqrt{x})^2-(2\sqrt{y})^2} \\\\= \dfrac{3\sqrt{x}(\sqrt{x}+2\sqrt{y})}{x-4y} .\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{3\sqrt{x}(\sqrt{x})+3\sqrt{x}(2\sqrt{y})}{x-4y} \\\\= \dfrac{3(\sqrt{x})^2+3(2)\sqrt{x}(\sqrt{y})}{x-4y} \\\\= \dfrac{3x+6\sqrt{x}(\sqrt{y})}{x-4y} .\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} \dfrac{3x+6\sqrt{x(y)}}{x-4y} \\\\= \dfrac{3x+6\sqrt{xy}}{x-4y} .\end{array}