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 April 22nd, 2018, 09:38 AM #1 Newbie   Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 Simplify expression I'm supposed to simplify the expression below. I've been at it for two hours and got nowhere significant, even with the provided steps in the solution With a>0 Solution is: or Thank you for your help. Cheers! By the way, how do we use LaTeX here? Last edited by bongcloud; April 22nd, 2018 at 09:50 AM.
April 22nd, 2018, 09:52 AM   #2
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Quote:
 Originally Posted by bongcloud E=\sqrt((x+3a)/(4a)+\sqrt((x-a)/(a)))+\sqrt((x+3a)/(4a)-\sqrt((x-a)/(a)))

Seems to be same as simply: E = 2sqrt(x+3a)/(4a)

 April 22nd, 2018, 10:00 AM #3 Newbie   Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 Pardonne-moi, I was trying to LaTeX
 April 22nd, 2018, 10:17 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 You can use dollar signs as delimiters for LaTeX. Thanks from bongcloud
 April 22nd, 2018, 10:22 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Je te pardonne mon enfant! Yikes! Sorry, but I just became VERY busy... But we have someone here named Joppy who simply adores those: he'll be along any minute and make that simplification stand on its head and spit wooden nickels Thanks from Joppy and bongcloud
 April 23rd, 2018, 12:15 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 What is $\left(1 + \frac12\sqrt{\frac{x - a}{a}}\right)^2$? Thanks from greg1313
April 23rd, 2018, 03:23 AM   #7
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Quote:
 Originally Posted by skipjack What is $\left(1 + \frac12\sqrt{\frac{x - a}{a}}\right)^2$?
$1+\sqrt{\frac{x-a}{a}}+\frac{1}{4}\lvert{\frac{x-a}{a}}\rvert$?

 April 23rd, 2018, 04:53 AM #8 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 Without apsolute value, only $\displaystyle 1+\sqrt{\frac{x-a}{a}}+\frac{x-a}{4a}=\sqrt{\frac{x-a}{a}}+\frac{3a+x}{4a}$.
 April 23rd, 2018, 09:55 AM #9 Newbie   Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 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 $1+\frac{1}{2}\sqrt{\frac{x-a}{a}}+\lvert 1-\frac{1}{2}\sqrt{\frac{x-a}{a}}\rvert$ Here's the full solution that was provided: \begin{aligned} \begin{aligned} &\frac{x+3a}{4a}-\sqrt{\frac{x-a}{a}}\geq 0\iff(x-5a)^2\geq 0.\\ &\text{Now we have} \end{aligned}\\ \begin{aligned} E^2&=2\frac{x+3a}{4a}+2\sqrt{\left(\frac{x+3a}{4a} \right)^2-\frac{x-a}{a}}\\ &=\frac{x+3a}{2a}+2\sqrt{\left(\frac{x-5a}{4a}\right)^2}\\ &=\frac{x+3a}{2a}+2\left|\frac{x-5a}{4a}\right|\\ &=\frac{x+3a}{2a}+\frac{\lvert x-5a\rvert}{2a},\ \text{because}\ a>0. \end{aligned}\\ \begin{aligned} &\text{Then if}\\ &i)\ a\leq x <5a, \text{then}\ E^2=\frac{x+3a}{2a}+\frac{-(x-5a)}{2a}=4\implies E=2;\\ &ii)\ 5a\leq x, \text{then}\ E^2=\frac{x+3a}{2a}+\frac{x-5a}{2a}=\frac{x-a}{a}\implies E=\sqrt{\frac{x-a}{a}}.\\ &\text{So,} \end{aligned} \end{aligned} E=\left\{\begin{aligned} &2, &&\text{for}\ a\leq x<5a\\ &\sqrt{\frac{x-a}{a}}, &&\text{for}\ 5a\leq x. \end{aligned} \right.
April 23rd, 2018, 12:05 PM   #10
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Quote:
 Originally Posted by lua Without apsolute value, only $\displaystyle 1+\sqrt{\frac{x-a}{a}}+\frac{x-a}{4a}=\sqrt{\frac{x-a}{a}}+\frac{3a+x}{4a}$.
I've managed to solve it in a completely different way. It's a very simple problem. I was just being a Dumbo, making a rewriting mistake repeatedly at the same place when solving.

I've tried solving it using skipjack's grouping, but I can't seem to make it work. Is it solvable that way? And also, how did you obtain that group skipjack? Is it from experience or is there a method I could make use of?

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