Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Test: Chapter 10 - Page 695: 21



Work Step by Step

Squaring both sides, the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{2x}=\sqrt{x+1}+1 \\\\ \left(\sqrt{2x}\right)^2=\left(\sqrt{x+1}+1\right)^2 \\\\ 2x=(\sqrt{x+1})^2+2(\sqrt{x+1})(1)+(1)^2 \\\\ 2x=x+1+2\sqrt{x+1}+1 \\\\ 2x-x-1-1=2\sqrt{x+1} \\\\ x-2=2\sqrt{x+1} .\end{array} Squaring both sides again, the equation above is equivalent to \begin{array}{l}\require{cancel} x-2=2\sqrt{x+1} \\\\ (x-2)^2=(2\sqrt{x+1})^2 \\\\ (x)^2-2(x)(2)+(2)^2=4(x+1) \\\\ x^2-4x+4=4x+4 \\\\ x^2-4x-4x+4-4=0 \\\\ x^2-8x=0 .\end{array} Factoring the $GCF=x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2-8x=0 \\\\ x(x-8)=0 .\end{array} Equating each factor to zero and then solving the variable, then the solutions are $ x=\{0,8\} .$ If $x=0,$ then \begin{array}{l}\require{cancel} \sqrt{2x}=\sqrt{x+1}+1 \\\\ \sqrt{2(0)}=\sqrt{0+1}+1 \\\\ \sqrt{0}=\sqrt{1}+1 \\\\ 0=1+1 \\\\ 0=2 \text{ (FALSE)} .\end{array} Hence, only $ x=8 $ satisfies the original equation.
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.