Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 89

Answer

$\color{blue}{11+4\sqrt6}$

Work Step by Step

RECALL: $(a+b)^2=a^2+2ab+b^2$ Use the formula above with $a=\sqrt3$ and $b=\sqrt8$ to obtain: $(\sqrt3+\sqrt8)^2 \\=(\sqrt3)^2+2(\sqrt3)(\sqrt8) + (\sqrt8)^2 \\=3 + 2\sqrt{3\cdot8}+8 \\=11 + 2\sqrt{24}$ Factor the radicand so that one factor is a perfect square, then bring out the square root of the perfect square factor to obtain: $=11+2\sqrt{4(6)} \\=11+2(2)\sqrt6 \\=\color{blue}{11+4\sqrt6}$
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