Answer
$\left\{-4,7\right\}$
Work Step by Step
Multiplying by the $LCD=
x^2
,$ the given equation $
1-\dfrac{3}{x}-\dfrac{28}{x^2}=0
,$ is equivalent to
\begin{align*}\require{cancel}
x^2\left(1-\dfrac{3}{x}-\dfrac{28}{x^2}\right)&=(0)x^2
\\\\
x^2(1)+x(-3)+1(-28)&=0
\\
x^2-3x-28&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(x-7)(x+4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x-7=0 & x+4=0
\\
x=7 & x=-4
.\end{array}
Checking the solutions in the given equation results to
\begin{array}{l|r}
\text{If }x=7: & \text{If }x=-4:
\\\\
1-\dfrac{3}{7}-\dfrac{28}{7^2}\overset{?}=0 &
1-\dfrac{3}{-4}-\dfrac{28}{(-4)^2}\overset{?}=0
\\\\
1-\dfrac{3}{7}-\dfrac{28}{49}\overset{?}=0 &
1+\dfrac{3}{4}-\dfrac{28}{16}\overset{?}=0
\\\\
\dfrac{49}{49}-\dfrac{21}{49}-\dfrac{28}{49}\overset{?}=0 &
\dfrac{16}{16}+\dfrac{12}{16}-\dfrac{28}{16}\overset{?}=0
\\\\
\dfrac{49-21-28}{49}\overset{?}=0 &
\dfrac{16+12-28}{16}\overset{?}=0
\\\\
0\overset{\checkmark}=0 &
0\overset{\checkmark}=0
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
1-\dfrac{3}{x}-\dfrac{28}{x^2}=0
$ is $\left\{-4,7\right\}$.
