College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 43

Answer

$x=9$

Work Step by Step

We are given: $\sqrt{x}-\sqrt{x-5}=1$ $\sqrt{x}=1+\sqrt{x-5}$ We square both sides: $(\sqrt{x})^{2}=(1+\sqrt{x-5})^{2}$ $(\sqrt{x})^{2}=(1+\sqrt{x-5})(1+\sqrt{x-5})$ We distribute: $x=1+2\sqrt{x-5}+(x-5)$ And combine like terms: $0=-4+2\sqrt{x-5}$ $4=2\sqrt{x-5}$ $2=\sqrt{x-5}$ We square both sides again: $2^{2}=(\sqrt{x-5})^{2}$ $4=x-5$ And solve for $x$: $x=9$
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.