Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Review Exercises - Page 657: 1


{$\pm 8$}

Work Step by Step

Given: $2x^2-3=125$ This implies $2x^2-128=0$ Take the common terms out. $2(x^2-64)=0$ This can be written as: $(x^2-8^2)=0$ $(x+8)(x-8)=0$ Thus, $x=\pm 8 $ Hence, $x=${$\pm 8$}
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.