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I have the equation $x^2=x$.

If I divide $x$ from both sides I get $x=1$.

Yet clearly $x$ can also equal $0$.

What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?

I hope this isn't a stupid question because I feel dumb asking about something so basic.

EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.

Kyle Delaney
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5 Answers5

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When you "divided both sides by $x$", you tacitly assumed that $x \neq 0$. It does not make sense to divide by zero.

In general, when confronted with problems like this, you can try to substitute in your solution from the beginning and go through the steps to see what goes awry. Here, starting with $x = 0$, the equation $x^2 = x$ is $0 = 0$. The next step is to divide both sides by $x$... whoops!

davidlowryduda
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  • Starting with $x = 0$, the equation $x^2 = x$ becomes $0 = 0$, right? Why would you divide either side by $x$ after that? – Kyle Delaney Aug 16 '15 at 02:31
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    @KyleDelaney: That is what you did, and he is showing you why it is wrong. – user21820 Aug 16 '15 at 05:17
  • Okay, this is a pretty good answer, except it doesn't explain how to make sure I don't miss one of the two possible solutions. I can see that there's a problem if I start with $x=0$, but the problem is that I originally missed that solution and so never thought to check what happens if $x=0$.

    It looks like my answer can be found here: http://math.stackexchange.com/questions/67994/why-should-you-never-divide-both-sides-by-a-variable-when-solving-an-equation

    I guess I'm just not supposed to divide by $x$ in the first place. Or rather, I'm supposed to take cases if I do, as chhro said.

     – Kyle Delaney Aug 18 '15 at 16:33
  • Okay, I actually still don't have a satisfactory answer to this. I've been studying calculus and I see teaching resources divide by variables all the time. I still have no idea how to discern when it's okay to do that. – Kyle Delaney Sep 07 '15 at 03:43
  • @KyleDelaney Any time you do anything to both sides of an equation, you should check what assumptions you are using. If you divide both sides by a variable, you are tacitly assuming that the variable is not $0$ --- otherwise, you've just committed an error. Similarly, if you take square roots, you are tacitly assuming that each side is nonnegative. You also should pay attention in the opposite direction: you can perform non-reversible operations that lose information. Multiplying by $0$ tells you nothing. Squaring can make you lose signs, e.g. although $(-1)^2 = 1^2$, we don't have $-1 = 1$. – davidlowryduda Sep 07 '15 at 06:24
  • The short answer is that you have mislearned the idea that you can "always divide out by $x$". This is not true, and $0$ is the only exception. – davidlowryduda Sep 07 '15 at 06:24
  • Okay, so when I see teachers dividing both sides by $x$ without taking extra steps to ensure that $x$ isn't $0$, are they doing it wrong? Or are they probably just leaving out the extra steps because they don't want to complicate whatever they're talking about? – Kyle Delaney Sep 08 '15 at 22:39
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$x^2=x\Rightarrow x^2-x=0\Rightarrow x(x-1)=0\Rightarrow $either $x=0$ or $x=1$

mathspuzzle
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  • Thanks, but that looks like the same thing Harish was saying. – Kyle Delaney Aug 16 '15 at 02:25
  • And neither of you answered the question. How would you know you're supposed to factor it instead of just dividing both sides by $x$, which is a perfectly valid step? – Kyle Delaney Aug 16 '15 at 02:34
  • @KyleDelaney: No it's not the same thing as what Harish said. "either X or Y" is not the same as "X and Y" so mathpuzzle's presentation is the correct one. See my other comments about why dividing by $x$ just like that is a totally invalid step. – user21820 Aug 16 '15 at 05:23
  • But when you factor $x^2-x$ into $x(x-1)$, aren't you dividing by $x$ in a way? Because to get one factor, you have to divide by the other factor. So factoring $x$ out of $x^2-x$ means dividing $x^2-x$ by $x$ and then expressing the two factors being multiplied together, right? – Kyle Delaney Aug 17 '15 at 02:52
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    @KyleDelaney No. Just no. Factoring is undoing a multiplication (which is uncritical) and does not involve a division (which would be critical). – GDumphart Aug 17 '15 at 07:39
  • @KyleDelaney: Do you agree that $2 \times 3 = 6$? Then whenever I see $6$ I can write in its place $2 \times 3$. Now do you agree that $x \times (x-1) = x^2 - x$ for any real $x$? Then whenever you see $x^2 - x$ you can ... – user21820 Aug 18 '15 at 14:13
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The given quadratic equation $x^2=x$ will have two real roots given as follows $$x^2-x=0$$ $$x(x-1)=0$$ $$x=0\ \ \ \ \textrm{or}\ \ \ x-1=0$$ $$x=0\ \ \ \ \textrm{or}\ \ \ x=1$$ Hence, we get $x=0$ or $x=1$

AJY
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Harish Chandra Rajpoot
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You cannot divide by $x$ if $x=0$ in the first place. That's why if you want to divide by $x$, you take cases. If $x=0$, we have a solution. If $x\neq0$, then we get $1$.

chhro
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  • So do you just always have to keep in mind that you should check to see if x could be 0? – Kyle Delaney Aug 16 '15 at 02:29
  • @KyleDelaney: No. You do not check if $x$ can be $0$. You check that you don't do any illegal thing, such as dividing by zero. You did not make sure of that when you divided by $x$. – user21820 Aug 16 '15 at 05:18
  • So are you saying that when I check that I don't do any illegal thing, I'd discover that $x$ could be $0$ because I'd have to take that as one of my cases? – Kyle Delaney Aug 17 '15 at 02:47
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Your methodology isn't wrong. You just need to modify it a bit. Before you divide both sides by $x$, you must check what happens when $x = 0$; that is, you must check whether or not $x=0$ is a solution. Once you have taken care of that, you are free to consider what happens when $x \ne 0$. In particular, now you can divide both sides by $x$.

(1) $x^2 = x$

(2) Clearly $x = 0$ is a solution.

(3) Now supose that $x \ne 0$. Then we can multiply both sides by $\dfrac 1x$.

(4) We get $x = 1$.

(5) The solution set is $x \in \{0, 1\}$.

Steven Alexis Gregory
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