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: 50



Work Step by Step

We are given: $\sqrt{2x}=\sqrt{3x+12}-2$ We square both sides: $(\sqrt{2x})^{2}=(\sqrt{3x+12}-2)^{2}$ $(\sqrt{2x})(\sqrt{2x})=(\sqrt{3x+12}-2)(\sqrt{3x+12}-2)$ $2x=3x+12-4\sqrt{3x+12}+4$ $2x-3x-16=-4\sqrt{3x+12}$ $-x-16=-4\sqrt{3x+12}$ $4\sqrt{3x+12}=x+16$ We square both sides again: $(4\sqrt{3x+12})^{2}=(x+16)^{2}$ $16\ (3x+12)=x^{2}+32x+256$ $48x+192=x^{2}+32x+256$ $x^{2}-16x+64=0$ And factor: $(x-8)^{2}=0$ Use the zero-factor property by equating each factor to zero: $x-8=0$ $x=8$
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.