## Intermediate Algebra (12th Edition)

$\dfrac{x-2\sqrt{xy}+y}{x-y}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}} ,$ 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{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}} \\\\= \dfrac{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})} \\\\= \dfrac{(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\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{(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x})^2-(\sqrt{y})^2} \\\\= \dfrac{(\sqrt{x}-\sqrt{y})^2}{x-y} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(\sqrt{x})^2-2(\sqrt{x})(\sqrt{y})+(\sqrt{y})^2}{x-y} \\\\= \dfrac{x-2(\sqrt{x})(\sqrt{y})+y}{x-y} .\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{x-2\sqrt{x(y)}+y}{x-y} \\\\= \dfrac{x-2\sqrt{xy}+y}{x-y} .\end{array}