Precalculus: Mathematics for Calculus, 7th Edition

$x=\pm3$
$x^{4}-8x^{2}-9=0$ Let $x^{2}$ be equal to $u$ If $u=x^{2}$, then $u^{2}=x^{4}$. Rewrite the original equation using the new variable $u$: $u^{2}-8u-9=0$ Solve by factoring: $(u-9)(u+1)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-9=0$ $u=9$ $u+1=0$ $u=-1$ Substitute $u$ back to $x^{2}$ and solve for $x$: $u=9$ $x^{2}=9$ $x=\sqrt{9}$ $x=\pm3$ $u=-1$ $x^{2}=-1$ $x=\sqrt{-1}$ Since $x=\sqrt{-1}$ won't yield real solutions, the final answer is $x=\pm3$