## Intermediate Algebra (12th Edition)

$\dfrac{a+2\sqrt{ab}+b}{a-b}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} ,$ 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{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}} \\\\= \dfrac{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})} \\\\= \dfrac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})} .\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{a}+\sqrt{b})^2}{(\sqrt{a})^2-(\sqrt{b})^2} \\\\= \dfrac{(\sqrt{a}+\sqrt{b})^2}{a-b} .\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{a})^2+2(\sqrt{a})(\sqrt{b})+(\sqrt{b})^2}{a-b} \\\\= \dfrac{a+2(\sqrt{a})(\sqrt{b})+b}{a-b} .\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{a+2\sqrt{a(b)}+b}{a-b} \\\\= \dfrac{a+2\sqrt{ab}+b}{a-b} .\end{array}