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I ran into this problem in the review section of my math text but I'm not sure how to go about solving it.

$$\frac{2x(x^2-9) - x^2(2x)}{(x^2-9)^2} = 0$$

I can't find a way to cancel the numerator with $(x^2 - 9)$. I guess I could multiply both sides of the equation by $(x^2 - 9)^2$ but I figured that wouldn't be allowed because so much of the expression would be 'lost'. Am I wrong? Could somebody point me in the right direction?

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  • $\begingroup$ One should keep track of any restrictions on the value of $x$ due to assumptions such as $(x^2-9)^2 \ne 0$. It turns out not to matter in the end, in this case, but sometimes it does matter. $\endgroup$
    – David K
    Jun 4, 2014 at 20:10

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Nothing will be lost, in fact, you would have then simplified your problem a lot. It is worth noting that if $Q(x)$ and $P(x)$ are two polynomials such that $Q(x)\neq0$ then $$\frac{P(x)}{Q(x)}=0\iff P(x)=0.$$ Therefore your problem becomes: $$\frac{2x(x^2-9) - x^2(2x)}{(x^2-9)^2} = 0\iff 2x(x^2-9) - x^2(2x)=0 \ \ \underline{\underline{\text{and}}} \ \ (x^2-9)^2\neq0.$$ How to move forward? Well, note that you have a common factor which is $2x$, then...

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$\frac{2x(x^2-9)-x^2(2x)}{(x^2-9)^2}=0$ if and only if $2x(x^2-9)-x^2(2x)=0$ and $x\ne\pm 3$. Solve.

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Your instincts are sound. In general there's nothing wrong with multiplying both sides of an equation by the same number. As for things going "lost," another word for that in this case is "simplifying the problem"! (EDIT: Now, if you wanted to multiply both sides of the equation by zero...that would be a case of lost information in a bad sense. But I think you know that.)

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