how do I solve this quadratic equation with a fraction?

Question

I seem to have trouble with quadratic equations when it comes to fractions and square roots.

$$ \frac{1}{x}+2x=3 $$ How do I solve this equation?

Answer 1

As well mentioned by @dxiv in the comment, you can easily see that $x$ cannot be zero (otherwise in the expression on the left we will do something which is not permitted to do so [which I shall let you find]). So, you can multiply the equation by $x$.

On multiplying whole equation by $x$, you get $1+2x^{2}=3x$ $\implies 2x^2-3x+1=0$. On factorising, it becomes $(2x-1)(x-1)$. So, $x=1/2$ or $x=1$. As required.

Answer 2

Or, multiply by $\, y = x^{-1}$ to get $\ y^2-3y +2 = (y\!-\!2)(y\!-\!1)= 0\ $ so $\, x^{-1} = y = 2,1$