## Intermediate Algebra (12th Edition)

$3\sqrt[]{5}-6$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{3}{\sqrt[]{5}+2} ,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products and the laws of radicals 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{3}{\sqrt[]{5}+2}\cdot\dfrac{\sqrt[]{5}-2}{\sqrt[]{5}-2} \\\\= \dfrac{3(\sqrt[]{5}-2)}{(\sqrt[]{5}+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{3(\sqrt[]{5}-2)}{(\sqrt[]{5})^2-(2)^2} \\\\= \dfrac{3(\sqrt[]{5}-2)}{5-4} \\\\= \dfrac{3(\sqrt[]{5}-2)}{1} \\\\= 3(\sqrt[]{5}-2) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3(\sqrt[]{5})+3(-2) \\\\= 3\sqrt[]{5}-6 .\end{array}