## Intermediate Algebra (12th Edition)

$-3+2\sqrt{2}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{1-\sqrt{2}}{1+\sqrt{2}} ,$ 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 by the conjugate of the denominator results to \begin{array}{l}\require{cancel} \dfrac{1-\sqrt{2}}{1+\sqrt{2}}\cdot\dfrac{1-\sqrt{2}}{1-\sqrt{2}} \\\\= \dfrac{(1-\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} \\\\= \dfrac{(1-\sqrt{2})^2}{(1+\sqrt{2})(1-\sqrt{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{(1-\sqrt{2})^2}{(1)^2-(\sqrt{2})^2} \\\\= \dfrac{(1-\sqrt{2})^2}{1-2} \\\\= \dfrac{(1-\sqrt{2})^2}{-1} \\\\= -(1-\sqrt{2})^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -[(1)^2-2(1)(\sqrt{2})+(\sqrt{2})^2] \\\\= -[1-2\sqrt{2}+2] \\\\= -[3-2\sqrt{2}] \\\\= -3+2\sqrt{2} .\end{array}