Find the roots of the following equation:
$x + \frac{1}{x} = 3, x \neq 0$

(N/A) Given the equation $x + \frac{1}{x} = 3$.
Multiplying throughout by $x$,we get:
$x^2 + 1 = 3x$
Rearranging the terms,we get the quadratic equation:
$x^2 - 3x + 1 = 0$
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 1, b = -3, c = 1$.
The discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$.
Since $D > 0$,the roots are real and distinct.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:
$x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.
Thus,the roots are $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$.