## Intermediate Algebra (12th Edition)

$2\sqrt{3}+\sqrt{10}-3\sqrt{2}-\sqrt{15}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} ,$ 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 of the given expression by the conjugate of the denominator, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} \cdot\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} \\\\= \dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})} .\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{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{(\sqrt{6})^2-(\sqrt{5})^2} \\\\= \dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{6-5} \\\\= \dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{1} \\\\= (\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5}) .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt{2}(\sqrt{6})+\sqrt{2}(\sqrt{5})-\sqrt{3}(\sqrt{6})-\sqrt{3}(\sqrt{5}) .\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} \sqrt{2(6)}+\sqrt{2(5)}-\sqrt{3(6)}-\sqrt{3(5)} \\\\= \sqrt{12}+\sqrt{10}-\sqrt{18}-\sqrt{15} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the index and extracting the root of that factor result to \begin{array}{l}\require{cancel} \sqrt{4\cdot3}+\sqrt{10}-\sqrt{9\cdot2}-\sqrt{15} \\\\= \sqrt{(2)^2\cdot3}+\sqrt{10}-\sqrt{(3)^2\cdot2}-\sqrt{15} \\\\= 2\sqrt{3}+\sqrt{10}-3\sqrt{2}-\sqrt{15} .\end{array}