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The area of a rectangle is $x^2 + 4x - 12$. What is the length and width of the rectangle?

The solution says the main idea is to factor $x^2 + 4x -12$.

So, since $-12 = -2 \times 6$ and $-2 + 6 = 4$, it can be written as $x^2 + 4x - 12 = (x - 2)(x + 6)$ since the length is usually the longer value, the length is $6$ and the width is $-2$.

I don't understand the logic to this solution at all. I understand $\text{length} \times \text{width} = \text{area}$, but outside of this information I don't understand how they got to this solution from the given information in the problem.

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    $\begingroup$ Are you sure that's all the information they gave you? Are the width and length (and $x$) supposed to be integers? $\endgroup$ – mr_e_man Apr 23 '18 at 2:50
  • $\begingroup$ We just know that $\ell w = x^2+4x-12$, whatever in the world $x$ is supposed to be. $\endgroup$ – Ted Shifrin Apr 23 '18 at 2:54
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    $\begingroup$ Was the question actually about a rectangle of area $A = \ell w = (x+4)(x) = x^2 + 4x = 12$ and find $x$, then $\ell$ and $w$? If so then everything else makes sense: You start with $x^2 + 4x = 12$, move over the 12 to get $x^2 +4x -12 = 0$, factor it as you've done to $(x-2)(x+6) = 0$, and you have two possible solutions: $x = 2$ or $x=-6$. But because negative lengths and widths don't make much sense, you keep only the solution $x=2$, whereupon you conclude that $\ell = x+4 = 6$ and $w = x = 2$. $\endgroup$ – Iwillnotexist Idonotexist Apr 23 '18 at 7:24
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    $\begingroup$ It seems likely that you have omitted important information from the question. Was that exactly the problem statement? What exactly did the solution say? (Did it just say to factor, and said nothing about how to interpret the factors?) Was there some statement earlier that said something like, "In questions 17 through 34, all rectangles are assumed to have width and length of the form $x + a$ and $x + b$ where $x$ is a variable"? $\endgroup$ – David K Apr 23 '18 at 12:27
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    $\begingroup$ The way you get $x^2 + 4x - 12 = (x-2)(x+6) \rightarrow x = 2,$ [or] $x = -6$ is when you have that $x^2 + 4x - 12 = 0$, because then $x^2 + 4x - 12 = (x-2)(x+6) = 0$, and the only two possible solutions to that equation can be $x = 2$ or $x = -6$. In this context, that would be if your area was zero, which doesn't make sense. Typically implicit in word problems like this is the assumption that $l, w > 0$. $\endgroup$ – C. Helling Apr 23 '18 at 15:28
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That is incorrect. Many rectangles, with different lengths and widths, can have the same area. Example: $2\times3 = 1\times6 = \pi\times\frac6\pi$

Also, the area is a function of $x$; if $x$ is not given, then the area is not given!

And width can't be negative.

Where did you find this "solution"? It's very wrong.


EDIT1

Perhaps they just wanted you to factor the expression. Then the "answer" would have the length $(x+6)$ , and the width $(x-2)$ . But even this isn't unique; it could be $(2x+12)$ and $(\frac12 x-1)$ . Someone gave you a bad question.

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  • $\begingroup$ The solution is fine. The interpretation of it is the problem. The rectangle is clearly generalised to rely on some number $x$, this only means that the area is not defined to a single value. It certainly is given, since we are told what it is, in the very question you are answering! $\endgroup$ – Nij Apr 23 '18 at 7:23
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    $\begingroup$ "And width can't be negative." There are cases where negative width is meaningful (as well as cases where negative area is). $\endgroup$ – Acccumulation Apr 23 '18 at 17:00
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The length is $(x+6)$ and the width is $(x-2)$. The area is $(x+6)(x-2)=x^2+4x-12$

You may take this answer as the length is 6 units longer and the width is 2 units shorter than a give number $x$

For example, given $x=10$, you may get that the area is $10^2+4\times10-12=128$, or, the area is $(10+6)(10-2)=16\times8=128$

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    $\begingroup$ I still don’t see what in the question, for this example with prohibits a rectangle of area = 128, what prevents the rectangle from having dimensions 32 and 4, or any other positive numbers who’s product is 128. Is x suppose to represent something physically? $\endgroup$ – Prince M Apr 23 '18 at 19:05

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