Answer
The solutions are $\frac{1}{3}$ and $-\frac{1}{2}$.
Work Step by Step
The given equation can be written as:
$(x^{-1})^2-x^{-1}-6=0$
Let $u=x^{-1}$.
Rewrite the equation above using $u$ to obtain:
$u^2-u-6=0$
Factor the trinomial to obtain:
$(u-3)(u+2)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&u-3=0 &\text{or} &u+2=0
\\&u=3 &\text{or} &u=-2
\end{array}
Since $u=x^{-1}$, then:
\begin{array}{ccc}
\\&u=3 &\text{or} &u=-2
\\&x^{-1}=3 &\text{or} &x^{-1}=-2
\\&(x^{-1})^{-1}=3^{-1} &\text{or} &(x^{-1})^{-1}=(-2)^{-1}
\\&x=\frac{1}{3} &\text{or} &x=-\frac{1}{2}
\end{array}
Therefore, the solutions are $\frac{1}{3}$ and $-\frac{1}{2}$.
