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Solve for $x$:
$x+\frac{1}{x}=3, x≠0$
Given:
Given quadratic equation is $x+\frac{1}{x}=3, x≠0$.
To do:
We have to solve the given quadratic equation.
Solution:
$x+\frac{1}{x}=3, x≠0$
$\frac{x(x)+1}{x}=3$
$\frac{x^2+1}{x}=3$
$x^2+1=3(x)$ (On cross multiplication)
$x^2+1=3x$
$x^2-3x+1=0$
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=1, b=-3$ and $c=1$.
Therefore, the roots of the given equation are
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-3)\pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}$
$x=\frac{3\pm \sqrt{9-4}}{2}$
$x=\frac{3\pm \sqrt{5}}{2}$
$x=\frac{3+\sqrt5}{2}$ or $x=\frac{3-\sqrt5}{2}$
The values of $x$ are $\frac{3+\sqrt5}{2}$ and $\frac{3-\sqrt5}{2}$. 
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