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A student divides both sides by $x-4$ and lost a solution $x=4$. How could you explain to the student that they are not allowed to divide by $x-4$

Here is the problem:

$x(x-4)=x(x-4)(x-5)$

I am having a hard time putting this in words for some reason. We know if the student divides by $x-4$ on both sides, not only do they lose a solution but technically they are dividing by $0$.

Does anyone have any other explanations

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    $\begingroup$ Well, in the present problem, $x=4$ is obviously a solution, right? If you now devide by $x-4$, it suddenly isn't anymore. Isn't this explanation enough? $\endgroup$ – Friedrich Philipp Feb 27 '16 at 19:16
  • $\begingroup$ Tell him that if he doesn't already know what $x$ is then he doesn't know that $x-4 \ne 0$ and so he doesn't know whether he can divide by it or not . Tell him to write in complete sentences,not "point form", emphasizing that he must include every "$\implies$" and "$\iff$". $\endgroup$ – DanielWainfleet Feb 27 '16 at 20:45
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You can divide by $x-4$ provided $x\neq 4$. For $x=4$ you need an another case.

The thinking goes as follows: I can divide by a number not equal to zero. I divide by $x-4$. When $x-4\neq0$ I can do that. When $x-4=0$ I check if $x=4$ is a solution.


Eventually you can show them this method

$$x(x-4)=x(x-4)(x-5)$$ $$x(x-4)-x(x-4)(x-5)=0$$ $$x(x-4)(1-(x-5))=0$$ $$-x(x-4)(x-6)=0\implies x\in\{0,4,6\}$$

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They can divide by $x-4$. But the rest of their work following that division must include the restriction that $x \neq 4$:

$$x(x-4) = x(x-4)(x-5) \\ \implies x = x(x-5); x \neq 4 \\ \vdots $$

You lose a solution because $x=4$ is a solution (by inspection) and $x \neq 4$ is our restriction in this line of work.

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The actual process of solving equations is simply a chain of biconditionals.

For example, $3x^2 + 3x = 9x \Longleftrightarrow 3x^2 - 6x = 0 \Longleftrightarrow 3x(x - 2) = 0 \Longleftrightarrow x = 2, 0$

In your example, the "chain of biconditionals" is broken.

It is not the case that $x(x-4)=x(x-4)(x-5) \Longleftrightarrow x = x(x-5)$. This fails for $x=4$, because division by zero is not permitted.

For the biconditional to hold, we must alter as follows: $x(x-4)=x(x-4)(x-5) \land x \neq 4 \Longleftrightarrow x = x(x-5)$

For the same reason, squaring both sides of an equation often leads to imprecise results, because the biconditional does not hold (the square function is not injective).

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Before dividing by $x-4$, they should split the search for solutions into two parts: (i) $x-4 = 0$ and (ii) $x-4 \neq 0$.

In the first case, we quickly check that $x=4$ is a solution and add that to our bag.

In the second case, since $x \neq 4$ we can happily divide across by $x-4$ to get $x = x (x-5)$ and continue the search (keeping in mind that we now know that $x \neq 4$, not that it matters here).

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Reorganizing things to one side, you have:

$$x(x-4)-x(x-4)(x-5)=0$$

Factoring yields:

$$x(x-4)(1-x+5)=0$$

If you have $ab=0$ this is true if and only if $a=0$ or $b=0$.

More generally, if $a_1a_2a_3\dots a_n=0$ this is true if and only if at least one of $a_1,a_2,a_3,\dots$ are zero.

So, the expression has solutions $x=0$ or $(x-4)=0$ or $(1-x+5)=0$

Simplifying, we have solutions $0,4,6$

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