Answer
Solution set: $\{2,10\}$
Work Step by Step
$g(a)=3$
$\displaystyle \frac{a}{4}+\frac{5}{a}=3$
Exclude solutions which yield 0 in denominators:
$a\neq 0\qquad (*)$
Multiply the equation with the LCD=$4a$
$a^{2}+20=12a$
$a^{2}-12a+20=0$
... $(x^{2}+bx+c$ is factored by searching for factors of c whose sum is b).
... $-10$ and $-2$ are factors of 20 whose sum is -12
$(a-10)(a-2)=0$
By the zero product principle,
$a\in\{2,10\}, $ ... both values satisfy (*).
Solution set: $\{2,10\}$
