Answer
$\{-1,\frac{-2\sqrt 6}{9},1,\frac{2\sqrt 6}{9} \}$
Work Step by Step
$3x^{\frac{4}{3}}-5x^{\frac{2}{3}}+2 = 0$
$3(x^{\frac{2}{3}})^{2}-5x^{\frac{2}{3}}+2 = 0$
Substituting $x^{\frac{2}{3}}=t$
$3t^{2}-5t+2 = 0$
Using quadratic formula,
$t= \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
Substituting $a=3, b=-5, c=2$
$t= \frac{-(-5)±\sqrt ((-5)^{2}-4(3)(2))}{2(3)}$
$t= \frac{5±\sqrt (25-24)}{6}$
$t= \frac{5±\sqrt (1)}{6}$
$t= \frac{5±1}{6}$
$t= \frac{5+1}{6}$ or $t= \frac{5-1}{6}$
$t= \frac{6}{6}$ or $t= \frac{4}{6}$
$t= 1$ or $t= \frac{2}{3}$
Replacing $t= x^{\frac{2}{3}}$
$x^{\frac{2}{3}}= 1$ or $x^{\frac{2}{3}} = \frac{2}{3}$
Let $x^{\frac{2}{3}}= 1$
$(x^{\frac{2}{3}})^{\frac{3}{2}}= ±1^{\frac{3}{2}}$
$x=± 1$
$x= 1$ or $x=-1$
Let $x^{\frac{2}{3}} = \frac{2}{3}$
$(x^{\frac{2}{3}})^{\frac{3}{2}} =±( \frac{2}{3})^{\frac{3}{2}}$
$x =±( \frac{2}{3})^{\frac{3}{2}}$
$x =±( \frac{8}{27})^{\frac{1}{2}}$
$x =±( \frac{2\sqrt 2}{3\sqrt 3})$
Rationalizing the denominator,
$x =±( \frac{2\sqrt 2}{3\sqrt 3}) \times \frac{\sqrt 3}{\sqrt 3}$
$x =±( \frac{2\sqrt 6}{9}) $
$x =( \frac{2\sqrt 6}{9}) $ or $x =-( \frac{2\sqrt 6}{9}) $
Solution Set: $\{-1,\frac{-2\sqrt 6}{9},1,\frac{2\sqrt 6}{9} \}$
