Answer
$\{-2,-1,0,1\}$
Work Step by Step
Recall that $|a|=b \implies a=b$ or $a=-b$.
Thus, using the rule above yields:
$x^2+x-1=1\quad$ or $\quad x^2+x-1=-1$
Solve each equation.
Solve $x^2+x-1=1$:
Subtract $1$ from both sides.
$x^2+x-1-1=1-1$
Simplify.
$x^2+x-2=0$
Rewrite $x$ as $2x-x$.
$x^2+2x-x-2=0$
Group the terms.
$(x^2+2x)+(-x-2)=0$
Factor each group.
$x(x+2)-1(x+2)=0$
Factor out $(x+2)$.
$(x+2)(x-1)=0$
Use Zero-Product Property.
$x+2=0\quad$ or $\quad x-1=0$
Solve for $x$.
$x=-2$ or $x=1$
Solve $x^2+x-1=-1$:
Add $1$ to both sides.
$x^2+x-1+1=-1+1$
Simplify.
$x^2+x=0$
Factor out $x$.
$x(x+1)=0$
Use Zero-Product Property.
$x=0\quad $ or $\quad x+1=0$
Solve for $x$.
$x=0\quad $ or $\quad x=-1$
Hence, the solution set of the equation is $\{-2,-1,0,1\}$.
