Answer
$\left\{-\dfrac{1}{4},2\right\}$
Work Step by Step
Multiplying by the $LCD=
r^2
,$ the given equation $
4-\dfrac{7}{r}-\dfrac{2}{r^2}=0
,$ is equivalent to
\begin{align*}\require{cancel}
r^2\left(4-\dfrac{7}{r}-\dfrac{2}{r^2}\right)&=(0)r^2
\\\\
r^2(4)+r(-7)+1(-2)&=0
\\
4r^2-7r-2&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(4r+1)(r-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
4r+1=0 & r-2=0
\\
4r=-1 & r=2
\\\\
r=-\dfrac{1}{4}
.\end{array}
Checking the solutions in the given equation results to
\begin{array}{l|r}
\text{If }r=-\dfrac{1}{4}: & \text{If }r=2:
\\\\
4-\dfrac{7}{-1/4}-\dfrac{2}{(-1/4)^2}\overset{?}=0 &
4-\dfrac{7}{2}-\dfrac{2}{2^2}\overset{?}=0
\\\\
4+\dfrac{7}{1/4}-\dfrac{2}{1/16}\overset{?}=0 &
4-\dfrac{7}{2}-\dfrac{2}{4}\overset{?}=0
\\\\
4+28-32\overset{?}=0 &
4+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\overset{?}=0
\\\\
0\overset{\checkmark}=0 &
4-4\overset{?}=0
\\
&
0\overset{\checkmark}=0
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
4-\dfrac{7}{r}-\dfrac{2}{r^2}=0
$ is $\left\{-\dfrac{1}{4},2\right\}$.
