## College Algebra 7th Edition

$x=\pm 3$
We solve by factoring: $x^{4}-8x^{2}-9=0$ $(x^{2}-9)(x^{2}+1)=0$ $(x-3)(x+3)(x^{2}+1)=0$ $x-3=0$ or $x+3=0$ or $x^2+1=0$ $x=3$ or $x=-3$ The last equation does not have a solution, since $x^2$ can not equal $-1$. Thus the solutions are $x=\pm 3$.