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 $(x=\frac{4}{3},\infty)$ ?
On
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
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 $k$”.
On
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.
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.