Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Review Exercises - Page 577: 60

Answer

$22-4\sqrt{30}$.

Work Step by Step

The given expression is $=(2\sqrt{3}-\sqrt{10})^2$ Use the special formula $(A-B)^2=A^2-2AB+B^2$ We have $A=2\sqrt3$ and $B=\sqrt{10}$. $=(2\sqrt3)^2-2(2\sqrt3)(\sqrt{10})+(\sqrt{10})^2$ Use product rule. $=12-4\sqrt{30}+10$ Simplify. $=22-4\sqrt{30}$.
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