Intermediate Algebra (12th Edition)

$\dfrac{\sqrt{x}+\sqrt{5}}{x-5}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{1}{\sqrt{x}-\sqrt{5}} ,$ 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 by the conjugate of the denominator results to \begin{array}{l}\require{cancel} \dfrac{1}{\sqrt{x}-\sqrt{5}}\cdot\dfrac{\sqrt{x}+\sqrt{5}}{\sqrt{x}+\sqrt{5}} \\\\= \dfrac{1(\sqrt{x}+\sqrt{5})}{(\sqrt{x})^2-(\sqrt{5})^2} \\\\= \dfrac{\sqrt{x}+\sqrt{5}}{(\sqrt{x})^2-(\sqrt{5})^2} .\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{5}}{x-5} .\end{array}