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: 53


$x=-\displaystyle \frac{2}{9}$ or $x=2$

Work Step by Step

We are given: $\sqrt{2\sqrt{7x+2}}=\sqrt{3x+2}$ We square both sides: $(\sqrt{2\sqrt{7x+2}})^{2}=(\sqrt{3x+2})^{2}$ $2\sqrt{7x+2}=3x+2$ And square both sides again: $(2\sqrt{7x+2})^{2}=(3x+2)^{2}$ $(2\sqrt{7x+2})(2\sqrt{7x+2})=(3x+2)(3x+2)$ $4\ (7x+2)=9x^{2}+12x+4$ And distribute: $28x+8=9x^{2}+12x+4$ $9x^{2}-16x-4=0$ And factor: $(9x+2)(x-2)=0$ Use the zero-factor property by equating each factor to zero: $(9x+2)=0$ or $(x-2)=0$ $x=-\displaystyle \frac{2}{9}$ or $x=2$
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.