April 24th, 2018, 08:30 AM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,617 Thanks: 2072 
Partly experience, in that it’s quite handy to have seen such a question before, but inspection would help as well. In your answer to my question, the absolute value isn’t needed. Consider also what you get if my expression has the “+” symbol replaced by a ““ symbol.

April 24th, 2018, 10:15 AM  #12  
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 
Do you mean $1+\sqrt{\frac{xa}{a}}+\frac{1}{4}\lvert{\frac{xa}{a}}\rvert$? Yes, I've realized that here: Quote:
However, it seems to me that we need to replace $\left(1\frac{1}{2}\sqrt{\frac{xa}{a}}\right)^2$ with $\left(\frac{1}{2}\sqrt{\frac{xa}{a}}1\right)^2 $. Then we have $ \left\lvert \frac{1}{2}\sqrt{\frac{xa}{a}}1\right\rvert=\left\{\begin{aligned} &\frac{1}{2}\sqrt{\frac{xa}{a}}1, &&\frac{1}{2}\sqrt{\frac{xa}{a}}1\geq 0\implies x\geq 5a\\ &1\frac{1}{2}\sqrt{\frac{xa}{a}}, &&\frac{1}{2}\sqrt{\frac{xa}{a}}1<0\implies a\leq x<5a \end{aligned} \right. $ Which produces the same solution set as in #9 $ E=\left\{\begin{aligned} &2, &&5a\geq x>a\\ &\sqrt{\frac{xa}{a}}, &&x\geq 5a \end{aligned} \right. $ Am I missing something? By the way, I appreciate your maieutic approach to teaching. Last edited by bongcloud; April 24th, 2018 at 10:40 AM.  

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