My Math Forum Simplify expression

 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 Display Modes Linear 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