## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$3x-2\sqrt{6x}+2$
Using $(a+b)^2=a^2+2ab+b^2$, or the square of a binomial, and the properties of radicals, the given expression, $(\sqrt{3x}-\sqrt{2})^2 ,$ is equivalent to \begin{array}{l}\require{cancel} (\sqrt{3x})^2+2(\sqrt{3x})(-\sqrt{2})+(-\sqrt{2})^2 \\\\= 3x-2\sqrt{6x}+2 .\end{array}