5
$\begingroup$

It's been a while since I worked with math, but I stumbled upon a rather 'simple' equation and the ways of solving it had me thinking for a bit.

Consider this equation:

$$x(1-x) = x(2 - \sqrt{1-x})$$

Any normal person would consider making the problem easier by dividing both left and right side of the equal sign by $x$, but bear in mind that one of the solutions might be $0$, so this is what makes me wonder, why would it be allowed to even divide by $x$ here, if one of its solutions might be $0$ then that is that $0/0$ is undefined... right?

Kindly refresh my memory.

$\endgroup$
5
  • 10
    $\begingroup$ As you say, dividing by $0$ is not allowed. So, split into two cases. $x=0$ is obviously a solution, and if $x\neq 0$, you can divide by it. (Note that in this case, it happens that there are no non-zero solutions, but that's not always the case). $\endgroup$
    – lulu
    Sep 6, 2022 at 20:20
  • 1
    $\begingroup$ Have a look into wheel theory. In that context, division is a unary operation and we define $\bot=0\times /0=0/0.$ $\endgroup$
    – Shaun
    Sep 6, 2022 at 20:21
  • 11
    $\begingroup$ That really isn't helpful here, @Shaun $\endgroup$ Sep 6, 2022 at 20:32
  • $\begingroup$ Division by $0$ CANNOT be made meaningful , neither with the "extended real line" nor with "wheel theory" or any other exotic method. This is not really a division anymore. But the main point is , correctly mentioned , that there is no reason to assume that the author has those exotic methods in mind. It is quite annoying that those comments almost always occur if any division-by-$0$ question is asked. $\endgroup$
    – Peter
    Sep 6, 2022 at 21:25
  • 1
    $\begingroup$ Using all caps doesn't make that rant any less false and misinformed. While bringing up the wheel of fractions or extended reals is out of context, saying that they don't give meaning to division by zero is immediately verifiably wrong. Moreover, $\endgroup$
    – plop
    Sep 6, 2022 at 22:24

2 Answers 2

11
$\begingroup$

You can divide both sides by $x$, but you must bear in mind that you are considering $x\neq 0$. Having said that, you can split the equation's solution into two steps. Firstly, consider $x = 0$ and notice if it is a solution or not (in the present case, it is). Secondly, you can consider that $x\neq 0$ in order to divide both sides by it so that you get the equation $1 - x = 2 - \sqrt{1 - x}$.

Additional Comments

With the purpose to solve such equation, there is no need to divide by $x$. Indeed, you can factor it as: \begin{align*} x(1 - x) = x(2 - \sqrt{1 - x}) & \Longleftrightarrow x(1 - x) - x(2 - \sqrt{1 - x}) = 0\\\\ & \Longleftrightarrow x(\sqrt{1 - x} - 1 - x) = 0 \end{align*}

Can you take it from here?

$\endgroup$
3
  • 6
    $\begingroup$ More generically, if $xy = xz$, then $x(y-z) = 0$, so $x=0$ or $y=z$. $\endgroup$
    – Dan
    Sep 6, 2022 at 20:40
  • 4
    $\begingroup$ Similar to what Dan is saying, the main principle here is that $ab = 0$ implies $a=0$ or $b=0$ for real (or complex) numbers $a,b$. That's why we would like to bring everything to one side of equation, factor it if possible and reduce the problem to two simpler equations, just like Átila did. $\endgroup$
    – Ennar
    Sep 6, 2022 at 20:50
  • 1
    $\begingroup$ I always like that factoring idea, which not only puts the factors on an equal footing (you could in fact arrange the equation to divide by either of the factors you end up with, it is the presentation which makes one division "obvious") but which also applies in more general contexts. $\endgroup$ Sep 6, 2022 at 21:13
3
$\begingroup$

Dividing by $x$ is permissible, but this reduces the equation into two cases where $x$ may or may not be $0$.

Let's go with a basic, almost trivial, example:

$$x^2 + x = x^3 - x^2$$

Now, at a glance, we can divide this by $x$, and get

$$x + 1 = x^2 - x$$

However, the caveats abound:

  • Yes, if $x=0$, we can't divide by it. We treat this case separately.
  • This is especially pertinent since $x=0$ is clearly a solution of the original.

So really we have that

$$x^2 + x = x^3 - x^2 \implies \begin{cases} x^2 + x = x^3 - x^2, & \text{if } x=0 \text{ (we'll see what happens here)} \\ x+1 = x^2 - x, & \text{if } x \ne 0\end{cases}$$

Well, the second case can be solved as usual via whatever means you prefer, e.g. quadratic formula after rearranging. This will give you some set of solutions (namely, $x = 1 \pm \sqrt{2}$).

The first equation is more simple: if $x=0$, the equation reduces to the true statement $0=0$. We conclude that $x=0$ is a solution. (If it resulted in something false, like $1=0$, then it would not be a solution.)

So the solution set of the original equation is the combination of both solution sets: $0, 1 + \sqrt 2, 1 - \sqrt 2$.


In your case, you divided by $x$.

If $x=0$, then you get something undefined in just $x/x$. BUT if you plug $x=0$ into the original equation you have, you get $0=0$: $0$ is thus a solution.

If $x \ne 0$, it becomes $1$. And then your equation simplifies and you can find some nonzero solutions.

$\endgroup$
2
  • $\begingroup$ The most trivial example is just $x^2=x.$ $\endgroup$ Sep 6, 2022 at 20:33
  • 1
    $\begingroup$ I know, but I wanted something a little less trivial than that to make things at least a little interesting. Being too trivial can be a problem in itself. $\endgroup$ Sep 6, 2022 at 20:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .