Answer
$\left\{\dfrac{1-i\sqrt{47}}{6},\dfrac{1+i\sqrt{47}}{6}\right\}$
Work Step by Step
Multiplying both sides by the $LCD=
r^2
,$ the given equation, $
\dfrac{4}{r^2}+3=\dfrac{1}{r}
,$ is equivalent to
\begin{align*}
r^2\left(\dfrac{4}{r^2}+3\right)&=\left(\dfrac{1}{r}\right)r^2
\\\\
1(4)+r^2(3)&=1(r)
\\
4+3r^2&=r
\\
3r^2-r+4&=0
.\end{align*}
The equation above has
\begin{align*}
a=
3
,\text{ }b=
-1
,\text{ and }c=
4
.\end{align*}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
x&=
\dfrac{-(-1)\pm\sqrt{(-1)^2-4(3)(4)}}{2(3)}
\\\\&=
\dfrac{1\pm\sqrt{1-48}}{6}
\\\\&=
\dfrac{1\pm\sqrt{-47}}{6}
\\\\&=
\dfrac{1\pm\sqrt{47\cdot(-1)}}{6}
\\\\&=
\dfrac{1\pm\sqrt{47}\cdot\sqrt{-1}}{6}
\\\\&=
\dfrac{1\pm\sqrt{47}\cdot i}{6}
&(\text{use }i=\sqrt{-1})
\\\\&=
\dfrac{1\pm i\sqrt{47}}{6}
.\end{align*}
Hence, the solution set of the equation $
\dfrac{4}{r^2}+3=\dfrac{1}{r}
$ is $\left\{\dfrac{1-i\sqrt{47}}{6},\dfrac{1+i\sqrt{47}}{6}\right\}$.
