## College Algebra (6th Edition)

$\sqrt (13 + \sqrt 2 + \frac{7}{3 + \sqrt 2})$ ........(1) To simplify first rationalize $\frac{7}{3 + \sqrt 2}$ To rationalize the denominator multiply the numerator and denominator by the conjugate of $3 + \sqrt 2$. The conjugate of $3 + \sqrt 2$ is $3 - \sqrt 2$ $\frac{7}{3 + \sqrt 2} \times \frac{3 - \sqrt 2}{3 - \sqrt 2}$ $= \frac{7(3 - \sqrt 2)}{(3 + \sqrt 2)(3 - \sqrt 2)}$ $( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $(\sqrt a)^{2}$ - $(\sqrt b)^{2}$. Therefore, $( 3 + \sqrt 2)( 3 - \sqrt 2)$ = $(3^{2})-(\sqrt 2^{2})$. $= \frac{7(3 - \sqrt 2)}{(3^{2})-(\sqrt 2^{2})}$ $= \frac{7(3 - \sqrt 2)}{9-2}$ $= \frac{7(3 - \sqrt 2)}{7}$ $=3 - \sqrt 2$ Substitute this value in Equation (1) $\sqrt (13 + \sqrt 2 + 3 - \sqrt 2)$ $\sqrt (13 + 3 + \sqrt 2 - \sqrt 2)$ $\sqrt 16$ $= 4$ $= 4$