Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - P.2 - Exponents and Radicals - Exercises - Page 25: 56

Answer

$-\frac{1}{2\sqrt 7+6}$

Work Step by Step

$\frac{\sqrt 7-3}{4}=\frac{\sqrt 7-3}{4}\frac{\sqrt 7+3}{\sqrt 7+3}=\frac{\sqrt 7\sqrt 7+3\sqrt 7-3\sqrt 7-3\times3}{4\sqrt 7+4\times3}=\frac{(\sqrt 7)^2-9}{4\sqrt 7+12}=\frac{7-9}{4\sqrt 7+12}=\frac{-2}{2(2\sqrt 7+6)}=\frac{-1}{2\sqrt 7+6}=-\frac{1}{2\sqrt 7+6}$
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