Appendix A - Equations and Inequalities - Page 420: 38

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

$x^{2}-64=0$ $(x)^{2}-(8)^{2}=0$ $(x+8)(x-8)=0$ $(x+8)=0$ 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 {$\pm8$}