# how do I solve this quadratic equation with a fraction?

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?

• Hint: $x=0$ can not be a solution, so multiply the whole equation by $x$. – dxiv Dec 12 '16 at 18:21
• A general strategy is, as a first step, to clear any fractions by multiplying by the LCD: softschools.com/math/algebra/topics/… – Daniel R. Collins Dec 12 '16 at 18:25

## 2 Answers

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.

• Nice post; but You should say something about why you can multiply by x (that it's not zero). – Namaste Dec 12 '16 at 18:27
• You can multiply by x even when it is 0! It just doesn't help solve the equation because then the equation reduces to "0= 0". – user247327 Dec 12 '16 at 18:35
• Thanks @amWhy, I have edited it now. – I am Back Dec 12 '16 at 18:37
• @user247327 no, because you'd have division by $0$ first, and you can't just multiply that by $0$ expecting anything times zero equals zero. – Simply Beautiful Art Dec 12 '16 at 23:59

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