Answer
$x=\pm 4$ or $x=\pm 1$
Work Step by Step
$ x^{4}-17x^{2}+16=0\qquad$...substitute $x^{2}$ for $u$
$ u^{2}-17u+16=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-17,\ c=16$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-17,\ a$ for $1$ and $c$ for $16$.
$ u=\displaystyle \frac{-(-17)\pm\sqrt{(-17)^{2}-4\cdot(16)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$u=\displaystyle \frac{17\pm\sqrt{289-64}}{2}$
$u=\displaystyle \frac{17\pm\sqrt{225}}{2}$
$ u=\displaystyle \frac{17\pm 15}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{17+15}{2}=\frac{32}{2}=16$ or $u=\displaystyle \frac{17-15}{2}=\frac{2}{2}=1$
Bring back $x$.
$x^{2}=16$ or $x^{2}=1$
$x=\pm 4$ or $x=\pm 1$
