What are the solutions to: $x(x-3)=x^2 -4$? What are the solutions to: $x(x-3)=x^2 -4$ ?
Subtracting $x^2$ from both side results in the loss of the solution $(x=\infty)$.
My book says the solutions are $(x=\frac{4}{3},\infty)$
But is $(x=-\infty)$ another solution?
Is the answer $(x=\frac{4}{3},\pm\infty)$ or $(x=\frac{4}{3},\infty)$ ?
 A: The asymptotic behavior of any two monic polynomials of the same degree is the same as $x\to+\infty$ and $x\to-\infty$, therefore approaching "infinity" to either side of the real line doesn't make a difference, so the author may simply write $\infty$ to mean $\pm\infty$.
Because your comment suggests that the author never writes $+\infty$ or $-\infty$, this reasoning works quite well.  If this is the reasoning, you could write $(\frac{4}{3},\pm\infty)$.

Another context is that the author makes extensive use of the projectively extended real line, where there is only one infinity, the $\infty$, as opposed to having both $+\infty$ and $-\infty$, in which case writing the solution as $\pm\infty$ is not acceptable.

Without knowing which is true, sticking to $\infty$ is safe.
A: It makes little sense to say that this equation “has an infinite solution”, because it is a degree $1$ equation.
On the other hand there are different contexts where it can make sense.
Consider the parametric equation
\[
x(x-3)=kx^2-4
\]
If we set $y=x^2$, this becomes
\[
\begin{cases}
y=x^2 \\[4px]
y-3x=ky-4
\end{cases}
\]
which can be interpreted as the intersection of a parabola with a pencil of lines, in this case $x=\frac{(1-k)}{3}y+\frac{4}{3}$, centered at $(4/3,0)$. Each of these lines has two distinct intersections with the parabola, with the exception of the lines for


*

*$k=25/16$ (tangent)

*$k=1$ (vertical line)


In olden textbooks (at least in Italy), such problems were studied in painful details and the exceptional case 2 was often treated by saying that one solution becomes infinite.
This makes sense from a projective geometry point of view, but I find it just confusing. My feeling is that such “infinite solution” were introduced in order to avoid students consider the case $k=1$ as tangency (because the equation has a single solution).
By the way, this misses also another line, the one with “infinite $k$”.
A: Whoa, whoa, whoa, back up.
$$x(x - 3) = x^2 - 4$$
Simplifying the left side:
$$x^2 - 3x = x^2 - 4$$
Now subtracting $x^2$ from both sides:
$$-3x = -4$$
And dividing both sides by $-3$:
$$x = \frac{-4}{-3} = \frac{4}{3}.$$
Graphing should lead to the same conclusion: four thirds is the only solution. As $x$ goes to either infinity, the gap between $x^2 - 4$ and $x^2 - 3x$ grows wider, though going to positive infinity, $x^2 - 4$ is greater, while going to negative infinity, $x^2 - 3x$ is greater.
In this view, infinity is something that can never be reached, that is always just beyond our reach. But if we can deal with infinities as if they were just a bunch of other algebraic quantities, does it make sense that $\infty^2 - 3 \infty = \infty^2 - 4$?
And what is $(-\infty)^2$? My guess is that $(-\infty)^2 = \infty$. Then $(-\infty)^2 + 3 \infty = (-\infty)^2 - 4$. I'm getting hung up on $(-\infty)^2 + 3 \infty$; if you can make sense of that as $(-\infty)^2 + 3 \infty = \infty$, then congratulations, you have shown that $x = -\infty$ is also a valid answer. But with Secretary of Educatuon Betsy DeVos wrecking our education system, good luck explaining it to anyone else.
A: $$\begin{align}
x(x-3)&=x^2 -4 \\ \\
\iff x^2-3x &= x^2 - 4\\ \\\iff 3x-4&=0\\ \\ \iff 3x&=4 \\ \\ \iff x &= \frac 43
\end{align}$$
