Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 636: 42

{$-8,\dfrac{1}{8}$}

Plug $f(x)=2$ Therefore, we have $2x^{2/3}+3x^{1/3}=2$ This implies $(2x^{1/3}-1)(x^{1/3}+2)=0$ Now, $(2x^{1/3}-1)=0$ or, $(x^{1/3})^3=(\dfrac{1}{2})^3$ or, $x=\dfrac{1}{8}$ and $(x^{1/3}+2)=0$ or, $(x^{1/3})^3=(-2)^3$ or, $x=-8$ Solution set: $x=${$-8,\dfrac{1}{8}$}