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: 42


The answer is $x=\sqrt5(2\sqrt3+3)\div30$.

Work Step by Step

Given: $4\sqrt15/(1+\sqrt3)=(1+\sqrt3)/x$ By Cross multiplication we get, $4\sqrt15*x=(1+\sqrt3)*(1+\sqrt3)$ $4x\sqrt15=(1+\sqrt3)^{2}$ $4x\sqrt3*\sqrt5=(1^{2}+\sqrt3^{2}+2*1*\sqrt3)$ $4x\sqrt3*\sqrt5=(1+3+2\sqrt3)$ $x=(4+2\sqrt3)/4*\sqrt3*\sqrt5$ $x=(2+\sqrt3)/2\sqrt15$ (Dividing by 2 in the numerator and denominator) $x=(2+\sqrt3)/2\sqrt15*\sqrt15/\sqrt15$ (Multiply and divide by $\sqrt15$) $x=(2+\sqrt3)/2\sqrt15*\sqrt15/\sqrt15$ $x=(2+\sqrt3)*\sqrt15/(2*15)$ $x=(2\sqrt15+\sqrt3*\sqrt15)/30$ $x=(2*\sqrt3*\sqrt5+3*\sqrt5)/30$ $x=\sqrt5(2\sqrt3+3)\div30$
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.