## College Algebra (10th Edition)

$x=\pm \frac{1}{\sqrt{27}}=\pm\frac{\sqrt{3}}{9}$
We solve by factoring: $3x^{4/3}+5x^{2/3}-2=0$ $[3x^{2/3}-1][x^{2/3}+2]=0$ $[3x^{2/3}-1]=0$ or $[x^{2/3}+2]=0$ $x^{2/3}=\frac{1}{3}$ or $x^{2/3}=-2$ $x=\pm (\frac{1}{3})^{3/2}$ or $x=\pm (-2)^{3/2}$ $x=\pm \frac{1}{\sqrt{27}}$ or $x=\pm \sqrt{-8}$= not real $x=\pm \frac{1}{\sqrt{27}}=\pm \frac{\sqrt{27}}{27}=\pm \frac{\sqrt{9*3}}{27}=\pm\frac{3\sqrt{3}}{27}=\pm\frac{\sqrt{3}}{9}$