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Suppose I am trying to solve the equation $49r^2=9r$. I would go about solving it like this:

$$49r^2=9r$$ $$49r^2-9r=0$$ $$\frac{49r^2-9r}{r}=\frac{0}{r}=0$$ $$49r-9=0$$ $$49r=9$$ $$r=\frac{9}{49}$$

But the problem is, $0$ is also a solution. How can I solve this equation in a way that doesn't ignore the solution $0$? Or is this something that I just have to look out for when solving equations?

Bill
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    Factor the expression as $r(9-49r) = 0$. This will give you both of your solutions. – ETS1331 Sep 14 '19 at 21:38
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    Instead of dividing by the variable, factor. Thus $49r^2-9r = 0$ becomes $r(49r-9) = 0$, splitting into the two solutions $r = 0$ and $49r-9=0$. – Andrew Sep 14 '19 at 21:38
  • You can divide only when the denominator is guaranteed to be non-zero. You divided by $r$ which is not allowed since $r$ can be equal to $0$. Factoring $r$ instead, we get: $r(49r-9)=0$, so either $r=0$ or $49r-9=0 \implies r=\frac{9}{49}$. – Hussain-Alqatari Sep 14 '19 at 21:49

4 Answers4

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I'd move everything to one side:

$$49r^2 - 9r = 0$$

and factor:

$$r(49r-9)=0$$

and then apply the general fact

$$AB=0\iff A = 0 \lor B=0$$

so that $r=0$ or $49r-9=0$ where the latter gives $r=\frac{9}{49}$ as a second solution. This way we keep equivalences ($\iff$) between all the solving steps and we miss no solutions.

Dividing by $r$ implies you're already in a scenario where $r\neq 0$, so that's where you have to be careful.

Henno Brandsma
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The thing is, when you divide by $r$, you are assuming that it is different from zero. There is no problem at doing so, but after solving the resulting equation, you have to plug in such value in the original one in order to check if $0$ is a solution or not. In your case, it is.

user0102
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When you divided by r, you Implicitly “got rid” of the 0 solution. If you want to keep it you should take

$$9r(\frac{49r}{9} - 1)=0$$

$$r(\frac{49r}{9} - 1)=0$$

Which evidently would lead to $r=0$ or $r=\frac{9}{49}$

Colin Hicks
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Be simple! The cancellation law works for nonzero factors. Thus, you can go like this:

By the cancellation law, $$49\mkern1.5mu r^2=9r\iff\begin{cases}\begin{aligned}[t] 49\mkern1.5mu r&=9 \iff r=\smash{\frac{9}{49}}&\text{if }r\ne 0, \\&\text{ or}\\ r&=0.\end{aligned}\end{cases}$$

Bernard
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