Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - 10-3 Operations With Radical Expressions - Practice and Problem-Solving Exercises - Page 630: 43


The answer is : $x=(9+3\sqrt10+6\sqrt2+4\sqrt5)$.

Work Step by Step

Given: $(2+\sqrt2)/(2-\sqrt2)=x/(3+\sqrt10)$ $\sqrt2(\sqrt2+1)/\sqrt2(\sqrt2-1)=x/(3+\sqrt10)$ $(\sqrt2+1)/(\sqrt2-1)=x/(3+\sqrt10)$ By cross multiplication we get, $(\sqrt2+1)*(3+\sqrt10)=x*/(\sqrt2-1)$ $x*(\sqrt2-1)=(\sqrt2+1)*(3+\sqrt10)$ $x=((\sqrt2+1)*(3+\sqrt10))/(\sqrt2-1)$ Multiply and divide by ($\sqrt2+1$), $x=((\sqrt2+1)*(3+\sqrt10))/(\sqrt2-1)*(\sqrt2+1)/(\sqrt2+1)$ $x=(\sqrt2+1)*(3+\sqrt10)*(\sqrt2+1)/(\sqrt2-1)*(\sqrt2+1)$ $x=(\sqrt2+1)^{2}*(3+\sqrt10)/(\sqrt2^{2}-1^{2})$ $x=(2+1+2\sqrt2)*(3+\sqrt10)$ $x=(3+2\sqrt2)*(3+\sqrt10)$ $x=(3*3+3*\sqrt10+2\sqrt2*3+2\sqrt2*\sqrt10)$ $x=(9+3\sqrt10+6\sqrt2+2\sqrt20)$ $x=(9+3*\sqrt2*\sqrt5+6\sqrt2+2\sqrt4*\sqrt5)$ $x=(9+3\sqrt10+6\sqrt2+4\sqrt5)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.