Answer
$x=3$
Work Step by Step
Multiplying both sides of the given equation, $
\dfrac{x^2+8}{x}-1=\dfrac{2(x+4)}{x}
,$ by the $LCD=x$, then,
\begin{array}{l}\require{cancel}
x\left( \dfrac{x^2+8}{x}-1 \right)=\left( \dfrac{2(x+4)}{x} \right)x
\\\\
1(x^2+8)+x(-1)=2(x+4)(1)
\\\\
x^2+8-x=2x+8
\\\\
x^2+(-x-2x)+(8-8)=0
\\\\
x^2-3x=0
\\\\
x(x-3)=0
\\\\
x=\{ 0,3 \}
.\end{array}
Upon checking, only $
x=3
$ satisfies the original equation.
