Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 636: 41

{$\dfrac{-1}{3},\dfrac{-3}{2}$}

Plug $f(x)=2$ Therefore, we have $3(\dfrac{1}{x}+1)^2+5(\dfrac{1}{x}+1)=2$ This implies $[3(\dfrac{1}{x}+1)-1][(\dfrac{1}{x}+1)+2]=0$ Now, $3(\dfrac{1}{x}+1)-1=0$ or, $-2x=3$ or, $x=\dfrac{-3}{2}$ and $(\dfrac{1}{x}+1)+2=0$ or, $-3x=1$ or, $x=\dfrac{-1}{3}$ Solution set: $x=${$\dfrac{-1}{3},\dfrac{-3}{2}$}