College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 178: 26

Answer

$x=2,6$

Work Step by Step

Separate the square roots on two sides. Then square both sides and combine like terms. Put all the terms on one side and factor by trial and error. Then solve for $x$ with the values found. $\sqrt {2x-3}-\sqrt {x-2}=1$ $\sqrt {2x-3}=1+\sqrt {x-2}$ $(\sqrt {2x-3})^2=(1+\sqrt {x-2})^2$ $2x-3=1+2\sqrt {x-2}+x-2$ $x-2=2\sqrt {x-2}$ $(x-2)^2=(2\sqrt {x-2})^2$ $x^2-4x+4=4(x-2)$ $x^2-8x+12=0$ $(x-2)(x-6)=0$ $x=2,6$
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.