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April 24th, 2018, 08:30 AM   #11
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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.
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April 24th, 2018, 10:15 AM   #12
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Do you mean $1+\sqrt{\frac{x-a}{a}}+\frac{1}{4}\lvert{\frac{x-a}{a}}\rvert$? Yes, I've realized that here:
Originally Posted by bongcloud View Post
I see. If we assume solution is $\mathbb{R}$, then $x-a\geq 0 \implies x\geq a$
I don't see how to proceed from there, though. For example, if I get rid of the roots, I'm left with $1+\frac{1}{2}\sqrt{\frac{x-a}{a}}+\lvert 1-\frac{1}{2}\sqrt{\frac{x-a}{a}}\rvert$
While typing this, I've realized I'd been a victim of poor rewriting skills yet again.
However, it seems to me that we need to replace $\left(1-\frac{1}{2}\sqrt{\frac{x-a}{a}}\right)^2$ with $\left(\frac{1}{2}\sqrt{\frac{x-a}{a}}-1\right)^2

Then we have
\left\lvert \frac{1}{2}\sqrt{\frac{x-a}{a}}-1\right\rvert=\left\{\begin{aligned}
&\frac{1}{2}\sqrt{\frac{x-a}{a}}-1, &&\frac{1}{2}\sqrt{\frac{x-a}{a}}-1\geq 0\implies x\geq 5a\\
&1-\frac{1}{2}\sqrt{\frac{x-a}{a}}, &&\frac{1}{2}\sqrt{\frac{x-a}{a}}-1<0\implies a\leq x<5a

Which produces the same solution set as in #9

&2, &&5a\geq x>a\\
&\sqrt{\frac{x-a}{a}}, &&x\geq 5a

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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