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 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
April 22nd, 2018, 10:00 AM  #3 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 
Pardonnemoi, 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.

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 
April 23rd, 2018, 12:15 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,826 Thanks: 2160 
What is $\left(1 + \frac12\sqrt{\frac{x  a}{a}}\right)^2$?

April 23rd, 2018, 03:23 AM  #7 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  
April 23rd, 2018, 04:53 AM  #8 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Without apsolute value, only $\displaystyle 1+\sqrt{\frac{xa}{a}}+\frac{xa}{4a}=\sqrt{\frac{xa}{a}}+\frac{3a+x}{4a}$.

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 $xa\geq 0 \implies x\geq a$ I don't see how to proceed from there, though. For example, if I get rid of the roots, I'm left with $1+\frac{1}{2}\sqrt{\frac{xa}{a}}+\lvert 1\frac{1}{2}\sqrt{\frac{xa}{a}}\rvert$ Here's the full solution that was provided: $ \begin{aligned} \begin{aligned} &\frac{x+3a}{4a}\sqrt{\frac{xa}{a}}\geq 0\iff(x5a)^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{xa}{a}}\\ &=\frac{x+3a}{2a}+2\sqrt{\left(\frac{x5a}{4a}\right)^2}\\ &=\frac{x+3a}{2a}+2\left\frac{x5a}{4a}\right\\ &=\frac{x+3a}{2a}+\frac{\lvert x5a\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{(x5a)}{2a}=4\implies E=2;\\ &ii)\ 5a\leq x, \text{then}\ E^2=\frac{x+3a}{2a}+\frac{x5a}{2a}=\frac{xa}{a}\implies E=\sqrt{\frac{xa}{a}}.\\ &\text{So,} \end{aligned} \end{aligned} $ $ E=\left\{\begin{aligned} &2, &&\text{for}\ a\leq x<5a\\ &\sqrt{\frac{xa}{a}}, &&\text{for}\ 5a\leq x. \end{aligned} \right. $ 
April 23rd, 2018, 12:05 PM  #10  
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  Quote:
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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