square root of a square root

**fezz349** · September 9th 2009, 03:10 PM

g of g(x) = 1/ sqrt( sqrt( 1 / x+2) +2 )

which could simpify to

4throot(x+2)/ sqrt(3) if you combine the bottom into one function.

What i'm interested in is where the sqrt(3) as a denominator came from, and what happened to the two 1 values, did they cancel out.

**Jameson** · September 9th 2009, 03:30 PM

Originally Posted by fezz349

g of g(x) = 1/ sqrt( sqrt( 1 / x+2) +2 )

which could simpify to

4throot(x+2)/ sqrt(3) if you combine the bottom into one function.

What i'm interested in is where the sqrt(3) as a denominator came from, and what happened to the two 1 values, did they cancel out.

Where did you get that simplification? What is the point of the problem?

**fezz349** · September 9th 2009, 08:03 PM

my cousin gave me the simplified answer that he received. I proceeded to then try to solve the probem myself and could not receive the same answer he got. I actually couldn't come up with one at all. If that's the right answer i'm looking for a push in the right direction so I'll understand the thought process that's involved with this kind of question.

**Jameson** · September 9th 2009, 08:25 PM

$\frac{1}{\sqrt{\sqrt{\frac{1}{x+2}+2}}}=\frac{1}{\ sqrt{\sqrt{\frac{2x+5}{x+2}}}}=\frac{1}{\left( \frac{2x+5}{x+2} \right) ^{\frac{1}{4}}}$

Assuming I understood what you typed correctly, this is how I would go about starting to simplify this. You have g(g(x)) though. Are you supposed to find g(x)?

**fezz349** · September 9th 2009, 09:34 PM

no sir, the + 2 is not included on the inner square root, it's by itself under the bigger one. I put the correct equation into the correct equation without a doubt, it's just simplifying it.

**Jameson** · September 9th 2009, 10:18 PM

Originally Posted by fezz349

no sir, the + 2 is not included on the inner square root, it's by itself under the bigger one. I put the correct equation into the correct equation without a doubt, it's just simplifying it.

Ok, my mistake.

So it's $\frac{1}{\sqrt{\sqrt{\frac{1}{x+2}}+2}}$

This is equal to $\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}=\frac{1}{\ sqrt{\frac{2\sqrt{x+2}+1}{\sqrt{x+2}}}}$

Now apply the square root separately to the numerator and denominator and keep simplifying. That's the way I'd go about it.