# Solving quadratic equation for inverse variable

I'm reading through some lecture notes and they show a quadratic equation, which I will just write in the usual way as

$$ax^2+bx+c=0$$

The notes say that, even though that equation can be solved in the usual fashion, it's easier to solve the corresponding equation for u=1/x.

I'm not sure how to solve a quadratic equation for the inverse of the original variable.

Any help greatly appreciated.

$$ax^2+bx+c=0$$ implies $$a+\frac bx+\frac c{x^2}=0$$ so that $$cu^2+bu+a=0$$ for $$u=1/x$$.

I will give an example as answer.

suppose we have to solve $$\frac{1}{x^2}+\frac{1}{x}-2=0$$.

We make the substitution $$u=1/x$$ then our equation become $$u^2+u-2=0$$. Solving for $$u$$ we then obtain $$u=1$$ or $$u=-2$$. Now we are able to find $$x$$:

$$u=\frac{1}{x}=1\Rightarrow x=1$$

$$u=\frac{1}{x}=-2\Rightarrow x=-\frac{1}{2}$$