Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is

$x+\frac{1}{x}=3, \quad x \neq 0$

$\Rightarrow \frac{x^{2}+1}{x}=3$

$\Rightarrow x^{2}+1=3 x$

$\Rightarrow x^{2}-3 x+1=0$

This equation is of the form $a x^{2}+b x+c=0$, where $a=1, b=-3$ and $c=1$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-3)^{2}-4 \times 1 \times 1=9-4=5>0$

So, the given equation has real roots.

Now, $\sqrt{\bar{D}}=\sqrt{5}$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-3)+\sqrt{5}}{2 \times 1}=\frac{3+\sqrt{5}}{2}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-3)-\sqrt{5}}{2 \times 1}=\frac{3-\sqrt{5}}{2}$

Hence, $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the roots of the given equation.