Intermediate Algebra (12th Edition)

$-19+\sqrt{77}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(\sqrt{7}-\sqrt{11})(2\sqrt{7}+3\sqrt{11}) ,$ use FOIL and the properties of radicals. Then combine like terms. $\bf{\text{Solution Details:}}$ 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{7}(2\sqrt{7})+\sqrt{7}(3\sqrt{11})-\sqrt{11}(2\sqrt{7})-\sqrt{11}(3\sqrt{11}) \\\\= 1(2)(\sqrt{7})^2+1(3)\sqrt{7}(\sqrt{11})-1(2)\sqrt{11}(\sqrt{7})-1(3)(\sqrt{11})^2 \\\\= 2(7)+3\sqrt{7}(\sqrt{11})-2\sqrt{11}(\sqrt{7})-3(11) \\\\= 14+3\sqrt{7}(\sqrt{11})-2\sqrt{11}(\sqrt{7})-33 .\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} 14+3\sqrt{7(11)}-2\sqrt{11(7)}-33 \\\\= 14+3\sqrt{77}-2\sqrt{77}-33 .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (14-33)+(3\sqrt{77}-2\sqrt{77}) \\\\= -19+\sqrt{77} .\end{array}