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If I have the equality $(x + 1)(x - 1) = \dfrac{-n}{m}(x - 1)$, where $m \not = 0$ and $x \in \mathbb{Z}^+$, is it an invalid operation to divide both sides by $(x - 1)$? Since $(x - 1)$ has the potential to equal $0$ if $x = 1$, it seems like it should be an invalid operation?

I would greatly appreciate it if people could please take the time to clarify this.

  • 1
    You always can separe it in cases: "for $x=1$, we have that... " and "for $x\neq 1$, we have ... ". – 2017-02-16
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    @rdias Indeed, that is an option. – 2017-02-16

2 Answers 2

-1

@The Pointer

You have:

$$(x + 1)(x - 1) = \frac{(-n)}{(m)}(x - 1)$$

Subtracting

$$\frac{(-n)}{(m)}(x - 1)$$

On both sides:

$$(x + 1)(x - 1) - \frac{(-n)}{(m)}(x - 1) = 0$$

Now you put $(x - 1)$ on evidence:

$$(x - 1)[(x + 1) - \frac{(-n)}{(m)}] = 0$$ Now, this can happen only if :

1) $$(x - 1) = 0$$ which leads to $$x_1 = 1$$ 2) $$(x + 1) - \frac{(-n)}{(m)} = 0$$ So, in this case, add $$\frac{(-n)}{(m)}$$ to both sides, which gives $$(x + 1) = \frac{(-n)}{(m)}$$ Subtracting 1 on both sides leads to $$x_2 = \frac{(-n)}{(m)} - 1$$

-2

If you divide by (x - 1) you lose one root. Instead, subtract it on both sides and factr it as : (X-1)[(x+1)-(-n/m)] = 0

So now you have that (x-1) = 0 Or (X+1) - (-n/m) = o

Which gives the roots x1 = 1 and x2 = (-n/m) - 1

If you have divided you would only find x2

  • 0
    I think this is incorrect, but it could just be that you haven't shown your calculations. – 2017-02-16