Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-3 Operations with Radial Expressions - Lesson Check - Page 616: 5

Answer

$3\sqrt5-\sqrt10$

Work Step by Step

$\frac{7\sqrt5}{3+\sqrt2}\longrightarrow$ multiply numerator and denominator by the conjugate of the denominator =$\frac{7\sqrt5}{3+\sqrt2}\times\frac{3-\sqrt2}{3-\sqrt2}\longrightarrow$ multiply =$\frac{7\sqrt5(3-\sqrt2)}{9-2}\longrightarrow$ use the distributive property =$\frac{21\sqrt5-7\sqrt{10}}{7}\longrightarrow$ divide by 7 =$3\sqrt5-\sqrt10$
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