Answer
$x=\displaystyle \pm\frac{\sqrt{2}}{3}$ or $x=\pm 2$
Work Step by Step
$ 9x^{4}-38x^{2}+8=0\qquad$...substitute $x^{2}$ for $u$
$ 9u^{2}-38u+8=0\qquad$... solve with the Quadractic formula. $a=9,\ b=-38,\ c=8$
$ u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-9,\ a$ for $4$ and $c$ for $5$.
$ u=\displaystyle \frac{-(-38)\pm\sqrt{(-38)^{2}-4\cdot(9)\cdot 8}}{2\cdot 9}\qquad$... simplify.
$u=\displaystyle \frac{38\pm\sqrt{1444-288}}{18}$
$u=\displaystyle \frac{38\pm\sqrt{1156}}{18}$
$ u=\displaystyle \frac{38\pm 34}{18}\qquad$... the symbol $\pm$ indicates two solutions.
$u=\displaystyle \frac{38+34}{18}=\frac{72}{18}=4$ or $u=\displaystyle \frac{38-34}{18}=\frac{4}{18}=\frac{2}{9}$
Bring back $x$.
$x^{2}=\displaystyle \frac{2}{9}$ or $x^{2}=4$
$x=\displaystyle \pm\frac{\sqrt{2}}{3}$ or $x=\pm 2$
