## Intermediate Algebra (12th Edition)

$\sqrt{3}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $\sqrt{\dfrac{12}{16}}+\sqrt{\dfrac{48}{64}} ,$ find a factor of the radicand that is a perfect power of the index. Then, extract the root of that factor. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Rewriting the radicand with a factor that is a perfect power of the index, the given expression is equivalent to \begin{array}{l}\require{cancel} \sqrt{\dfrac{4}{16}\cdot3}+\sqrt{\dfrac{16}{64}\cdot3} \\\\= \sqrt{\dfrac{1}{4}\cdot3}+\sqrt{\dfrac{1}{4}\cdot3} \\\\= \sqrt{\left( \dfrac{1}{2} \right)^2\cdot3}+\sqrt{\left(\dfrac{1}{2}\right)^2\cdot3} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{1}{2}\sqrt{3}+\dfrac{1}{2}\sqrt{3} \\\\= \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2} .\end{array} Combining the like radicals results to \begin{array}{l}\require{cancel} \dfrac{\sqrt{3}+\sqrt{3}}{2} \\\\= \dfrac{2\sqrt{3}}{2} \\\\= \dfrac{\cancel{2}\sqrt{3}}{\cancel{2}} \\\\= \sqrt{3} .\end{array}