College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises: 54

Answer

$x=\pm1$

Work Step by Step

$x^{8}+15x^{4}=16$ $x^{8}+15x^{4}-16=0$ $(x^{4}+16)(x^{4}-1)=0$ $x^{4}+16=0$ or $x^{4}-1=0$ $x^4=-16$ or $x^4=1$ $x=\pm\sqrt[4]{-16}$ or $x=\pm\sqrt[4]{1}=\pm1$ The left solution is not real, so we ignore it. Thus the solution is $x=\pm1$
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.