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April 22nd, 2018, 09:38 AM   #1
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Smile 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.
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April 22nd, 2018, 09:52 AM   #2
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Quote:
Originally Posted by bongcloud View Post
E=\sqrt((x+3a)/(4a)+\sqrt((x-a)/(a)))+\sqrt((x+3a)/(4a)-\sqrt((x-a)/(a)))
Whadda heck does that mean?

Seems to be same as simply: E = 2sqrt(x+3a)/(4a)
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April 22nd, 2018, 10:00 AM   #3
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Pardonne-moi, I was trying to LaTeX
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April 22nd, 2018, 10:17 AM   #4
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You can use dollar signs as delimiters for LaTeX.
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April 22nd, 2018, 10:22 AM   #5
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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
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April 23rd, 2018, 12:15 AM   #6
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What is $\left(1 + \frac12\sqrt{\frac{x - a}{a}}\right)^2$?
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April 23rd, 2018, 03:23 AM   #7
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Quote:
Originally Posted by skipjack View Post
What is $\left(1 + \frac12\sqrt{\frac{x - a}{a}}\right)^2$?
$1+\sqrt{\frac{x-a}{a}}+\frac{1}{4}\lvert{\frac{x-a}{a}}\rvert$?
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April 23rd, 2018, 04:53 AM   #8
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Without apsolute value, only $\displaystyle 1+\sqrt{\frac{x-a}{a}}+\frac{x-a}{4a}=\sqrt{\frac{x-a}{a}}+\frac{3a+x}{4a}$.
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April 23rd, 2018, 09:55 AM   #9
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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.
$
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April 23rd, 2018, 12:05 PM   #10
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Quote:
Originally Posted by lua View Post
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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