Infinite solution in quadratic equation In my textbook it says that some quadratic equations can have infinite solutions. For example if $x^2-2x=x^2-3$ then rather than cancelling $x^2$ from both sides the book lets $x=\frac{1}{m}$ and then rearranges to get $1-2m=1-3m^2$ or $m(3m-2)=0$ therefore $m=0$ or $m=\frac{2}{3}$ giving $x=\infty$ or $\frac{3}{2}$. This is the first time I see this so I am just wondering if $\infty$ is actually a valid solution?
 A: Think of the real number line as being bent to a circle and glued together at $+\infty$ and $-\infty$, yielding a single point called $\infty$ (i.e. you do not distinguish between $-\infty$ and $+\infty$). This is called compactification (in fact a particular one, it's not unique) and yields a set $\mathbb{R}\cup\{\infty\}$ that even allows to do some "unusual" arithmetic such as $a/0=\infty$ and $a/\infty=0$ (for $a \neq 0$). So in this sense, $\infty$ is indeed a valid solution.
A: Since $$\lim_{x\to\infty}x^2 \gg\lim_{x\to\infty}x\gg\lim_{x\to\infty}c$$
Here $c$ is some constant less than $\infty$.
A: My immediate response would be no.
There are many reasons for this, I'll propose just two:
Firstly, $m = 0$ with $x = \frac{1}{m}$ has no solutions for $x$. Division by zero is not allowed --- and, depending on your approach, $1/0$ could be any number you want. Just compute the limit from either direction to see at least two possible alternatives; from the positive side of $0$, $x/0$ approaches $+\infty$ but from the negative side of $0$, $x/0$ approaches $-\infty$.
Secondly, any polynomial can have a solution that is infinity in this case, if one abides by the proposed axioms of infinity ($\infty + c = \infty$, $k\times\infty = \infty$ for all $c, k \in \mathbb{R}$).
In higher levels of Mathematics it may be the case that you can have infinity as a solution, but I've always steered clear of it. Infinity does weird things!
