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This is probably a really easy task for the people of this site judging by what is normally discussed here. (I'm amazed by the knowledge here!)

I have this expression that I need to simplify:

$$\frac{(x-2)^2}{\sqrt{(2-x)^2}}$$

From what I've understood $\sqrt{a^2}$ is the same thing as $|a|$ so I can rewrite the expression as:

$$\frac{(x-2)^2}{|2-x|}$$

...and this is where I can't get any further. My guess is that I should think of the definition for absolute value and maybe do something like:

$$\left\{ \begin{array}{ccc} 2-x & if & x < 2 \\ -(2-x) & if & x > 2 \end{array} \right.$$

But I don't know... I don't understand it, I'm just guessing. Also, what about the numerator? When searching the net, it's like only the absolute value expression is the one you work with (in examples like the one above). I just don't get it. :(

I'd really be grateful for an explanation of this problem!

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You are doing fine up to $\dfrac{(x-2)^2}{|2-x|}$. I would prefer to express this as $\dfrac{(x-2)^2}{|x-2|}$, looks a bit nicer.

Now a little trick: $\dfrac{x-2}{|x-2|}=\dfrac{|x-2|}{x-2}$. So our expression becomes $\dfrac{(x-2)|x-2|}{x-2}$, which simplifies to $|x-2|$. (But the two expressions are not exactly equal, they are only equal when $x\ne 2$.)

There are endlessly many ways other ways to do the calculation. For example, in your starting expression, both top and bottom are positive. So you can rewrite it as $$\frac{\sqrt{(x-2)^4}}{\sqrt{(2-x)^2}}, \quad\text{which is}\quad \sqrt{\frac{(x-2)^4}{(2-x)^2}}.$$ There is some cancellation, since $(x-2)^2=(2-x)^2$ so we end up with $\sqrt{(x-2)^2}$, which is $|x-2|$.

Or else note that apart from some possible sign difficulties, your original expression should simplify to something like $x-2$. More formally, the absolute value of your expression is $|x-2|$, so your expression will simplify to something of shape $\pm|x-2|$, and we have to figure out where plus and where minus. But the expression is positive if $x\ne 2$, so the expression simplifies to $|x-2|$.

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  • $\begingroup$ Very nice. $\quad$ $\endgroup$ – Brian M. Scott Sep 8 '12 at 20:28
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You're doing well. First, notice that $|2-x| = |-(x-2)|= |x-2|$. So the expression we're working with is $\frac{(x-2)^2}{|x-2|}$. If $x > 2$, this is the same as $\frac{(x-2)^2}{x-2} = x-2$. If $x < 2$, this is $\frac{(x-2)^2}{-(x-2)} = -(x-2)$. This is the same result you got, and in fact there's a shorter way to write it: $|x-2|$.

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