Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Appendix A - Equations and Inequalities - Page 420: 38


{$-8,+8$} or {$\pm8$}

Work Step by Step

$x^{2}-64=0$ $(x)^{2}-(8)^{2}=0$ $(x+8)(x-8)=0$ $(x+8)=0$ or $(x-8)=0$ $x=-8$ or $x=8$ Check: Substituting $x=-8$ in the equation, $(-8)^{2}-(8)^{2}=0$ $64-64=0$ $0=0$ Substituting $x=8$ in the equation, $(8)^{2}-(8)^{2}=0$ $64-64=0$ $0=0$ Both values check and the solution set is {$-8,+8$} or {$\pm8$}
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