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: 90

Answer

$\color{blue}{15+10\sqrt2}$

Work Step by Step

RECALL: $(a+b)^2=a^2+2ab+b^2$ Use the formula above with $a=\sqrt5$ and $b=\sqrt{10}$ to obtain: $(\sqrt5+\sqrt{10})^2 \\=(\sqrt5)^2+2(\sqrt5)(\sqrt{10}) + (\sqrt{10})^2 \\=5 + 2\sqrt{5\cdot10}+10 \\=15 + 2\sqrt{50}$ Factor the radicand so that one factor is a perfect square, then bring out the square root of the perfect square factor to obtain: $=15+2\sqrt{25(2)} \\=15+2(5)\sqrt2 \\=\color{blue}{15+10\sqrt2}$
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