$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$ leading to contradictory solutions I have the following equation
$$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$$
And to find the critical values for the absolute value function I carried out polynomial long division and ended up with
$$1+ \frac{a}{x-2a}, x>2a$$
$$-(1+ \frac{a}{x-2a}), x<2a$$
I first attempted to solve this using the original form of both equations
For the case where $x>2a$:
$$\frac{(x-a)^2}{(x-2a)^2} =\frac{x-a}{x-2a} $$
$$\frac{x-a}{x-2a} =1$$
$$x-a=x-2a$$
However this lead me to the answer
$$-a=-2a$$
Which is self contradictory, I am extremely confused on where I messed up, have I made a problem cross multiplying or is it something else?
 A: If $$\frac{(x-a)^2}{(x-2a)^2}=\left|\frac{x-a}{x-2a} \right|$$
Let's $r = \left|\frac{x-a}{x-2a} \right|$.
This just says that $r^2=r$, hence $r=0$ or $r=1$.
If $r=0$, we have $x=a$.
Otherwise, $|x-a|=|x-2a|$, the distance from $a$ is equal to the distance from $2a$, hence $x=\frac{3a}{2}$.

Now, let's examine your solution.
If $x> 2a$, we know that $x -a > a$, it is unclear if $|x-a| = x-a$ isn't it?
In fact, if $x=a$, then you divided by $x-a$, which is $0$.
If $x=\frac32a$ and $\frac32 a > 2a$, then we have $a<0$. $x-a=\frac12a$ which is negative.
A: For the case you describe, you didn't mess anything up:  what you found tells you there is no solution for $ \ x > 2a \ $ .  (However, your characterization of the rational function after polynomial division is incomplete.)
In your original equation,
$$\frac{(x-a)^2}{(x-2a)^2} \ = \ \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \ \ , $$
the rational function within the absolute-value brackets has two "special points": $ \ x = 2a \ $ , where the function is undefined, and $ \ x = a \ $ , where the function equals zero.
The real numbers are divided into three intervals then, $ \ x < a \ , \ a < x < 2a \ , \ \text{and} \ x > 2a \ . $  In the first interval, the numerator and denominator are both negative, while in the third interval, both are positive.
So for these cases, $ \  \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \ = \ \frac{x-a}{x-2a} \ , $ making the original equation
$$ \left( \frac{x-a}{x-2a} \right)^2 \ = \ \frac{x-a}{x-2a}  \ \ , $$
which can only be true when both sides equal zero or $ \ 1 \ $ .  The former is the case, as remarked above, for $ \ x = a \ $ ; in the latter case, the numerator and denominator cannot be equal for any real number.  Hence, there are no solutions to the original equation in the intervals  $ \ x < a \ \  \text{or} \ \ x > 2a \ . $
In the interval $ \ a < x < 2a \ , $ however, the numerator is positive while the denominator is negative, so we have
$$  \left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert \  = \ -\frac{x-a}{x-2a} \ = \ \frac{x-a}{2a-x} \ \ \rightarrow \ \ \frac{(x-a)^2}{(x-2a)^2} \ = \  \frac{x-a}{2a-x} \ \ . $$
Since $ \ x \ne a \ , \ x \ne \ 2a \ , $ it is "safe" to cross-multiply and divide out factors, leading to
$$ (x-a)^2 · (2a-x) \ = \ (x-a) · (x-2a)^2 \ \ \Rightarrow \ \ (x-a)  · ( -1) \ = \  (x-2a)  $$
$$ \Rightarrow \ \ (x-2a) \ + \ (x-a) \ = \ 0 \ \ \Rightarrow \ \ 2x - 3a \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac32 a \ \ . $$
The original equation thus has two solutions, $ \ x = a \ $ and $ \ x \ = \ \frac32 a \ \ . $
