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 - Summary Exercises on Solving Equations - Page 136: 5

Answer

No solution.

Work Step by Step

We solve: $\sqrt{x+2}+5=\sqrt{x+15}$ First we square both sides: $(\sqrt{x+2}+5)^{2}=(\sqrt{x+15})^{2}$ $(\sqrt{x+2}+5)(\sqrt{x+2}+5)=(\sqrt{x+15})(\sqrt{x+15})$ $x+2+10\sqrt{x+2}+25=x+15$ $x+27+10\sqrt{x+2}=x+15$ $10\sqrt{x+2}=15-27$ $10\sqrt{x+2}=-12$ $5\sqrt{x+2}=-6$ We square both sides again: $(5\sqrt{x+2})^{2}=(-6)^{2}$ $25(x+2)=36$ $25x+50=36$ $25x=-14$ $x=-\frac{14}{25}$ However, the solution $x=-\frac{14}{25}$ does not work in the original equation: $\sqrt{-\frac{14}{25}+2}+5=\sqrt{-\frac{14}{25}+15}$ $\sqrt{-\frac{14}{25}+\frac{50}{25}}+5=\sqrt{-\frac{14}{25}+\frac{375}{25}}$ $\sqrt{\frac{36}{25}}+5=\sqrt{\frac{361}{25}}$ $\frac{6}{5}+\frac{25}{5}=\frac{19}{5}$ $\frac{31}{5}=\frac{19}{5}$ $31=19$ This is a false statement, so there is no solution.
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.