College Algebra 7th Edition

Published by Brooks Cole

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 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 $x=\pm1$

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