User Name Remember Me? Password

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 April 24th, 2018, 08:30 AM #11 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2207 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. Thanks from bongcloud April 24th, 2018, 10:15 AM   #12
Newbie

Joined: Apr 2018
From: Banovo Brdo

Posts: 12
Thanks: 0

Do you mean $1+\sqrt{\frac{x-a}{a}}+\frac{1}{4}\lvert{\frac{x-a}{a}}\rvert$? Yes, I've realized that here:
Quote:
 Originally Posted by bongcloud 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 \end{aligned} \right.

Which produces the same solution set as in #9

E=\left\{\begin{aligned} &2, &&5a\geq x>a\\ &\sqrt{\frac{x-a}{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. Tags expression, simplify Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Help123 Calculus 3 January 8th, 2018 03:02 PM Chikis Algebra 4 December 18th, 2015 02:18 AM jaredbeach Algebra 2 September 2nd, 2011 09:00 AM football Algebra 4 March 14th, 2011 08:48 AM jaredbeach Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      