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