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

 April 22nd, 2018, 10:38 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 10:50 AM. April 22nd, 2018, 10:52 AM   #2
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 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, 11:00 AM #3 Newbie   Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0 Pardonne-moi, I was trying to LaTeX  April 22nd, 2018, 11:17 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 You can use dollar signs as delimiters for LaTeX. Thanks from bongcloud April 22nd, 2018, 11:22 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 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, 01:15 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,131 Thanks: 2340 What is $\left(1 + \frac12\sqrt{\frac{x - a}{a}}\right)^2$? Thanks from greg1313 April 23rd, 2018, 04:23 AM   #7
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 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, 05:53 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, 10:55 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, 01:05 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? 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 04:02 PM Chikis Algebra 4 December 18th, 2015 03:18 AM jaredbeach Algebra 2 September 2nd, 2011 10:00 AM football Algebra 4 March 14th, 2011 09:48 AM jaredbeach Calculus 1 December 31st, 1969 04:00 PM

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