$$(x^2+6x+9)^{-\frac{1}{2}} \cdot (x+3)^2$$

The answer is $x+3$, but I don't understand how? There is no restriction, should it not be as follows?

$$\frac{1}{\sqrt{(x+3)^2}} \cdot (x+3)^2$$

$$\frac{1}{|x+3|} \cdot (x+3)^2$$

So shouldn't there be two answers,

$-(x+3)$ if $x \lt -3$ and $(x+3)$ if $x \ge -3$

How am I supposed to solve these problems? I am confused with the whole square root deal.

In order to remove confusion, I got two answers, but apparently there is only one?

  • 3
    $\begingroup$ I think you're correct here, and whoever's given the answer has made an error $\endgroup$ – Mathmo123 Aug 12 '14 at 19:06
  • $\begingroup$ This is completely accurate. (I would have made the same error though. :p ) $\endgroup$ – Guy Aug 12 '14 at 19:22
  • $\begingroup$ Shortly: $|x+3|$. $\endgroup$ – Yves Daoust Aug 12 '14 at 19:29

The two answers can be combined into $|x+3|.$ But you're right it shouldn't be just $x+3$ unless one has already assumed somewhere that $x>-3.$

Replace $(x+3)^2$ at your last step by the equivalent $(|x+3|)^2$ and cancel one copy with the denominator to the left, of course assuming $x \neq -3.$ This last restriction is imposed by the original expression as your previous step shows.


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