Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 67

$x=-\frac{1}{2}$ or $x=-2$

$\displaystyle \frac{1}{(x+1)^{2}}=\frac{1}{x+1}+2$ We multiply both sides by $(x+1)^{2}$: $1=(x+1)+2(x+1)^{2}$ $2(x+1)^{2}+(x+1)-1=0$ We solve by factoring: $[2(x+1)-1][(x+1)+1]=0$ $(2x+2-1)(x+2)=0$ $(2x+1)(x+2)=0$ $(2x+1)=0$ or $(x+2)=0$ $x=-\frac{1}{2}$ or $x=-2$