# Simplifying Square Root w/in square root

• Sep 6th 2006, 08:10 PM
tesla1410
Simplifying Square Root w/in square root
I've been asked to simplify and I don't know how.

1/[square root of 1/(square root of x+2) +2]

Wow.
• Sep 6th 2006, 10:26 PM
CaptainBlack
Quote:

Originally Posted by tesla1410
I've been asked to simplify and I don't know how.

1/[square root of 1/(square root of x+2) +2]

Wow.

Fist we need to get it clear what you have been asked to simplify, is it:

$
\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}\ ?
$

Whatever it truns out the exact problem is the key trick you will need
to employ is:

$\frac{1}{\sqrt{something}}=\frac{\sqrt{something}} {something}$

RonL
• Sep 7th 2006, 06:35 AM
tesla1410
Yes, you've stated the problem correctly. What's throwing me is the square root fraction w/in the square root - but then you probably know that's the hang up.

BTW, is there some way I can use the trick symbols that so many others use? It would make writing a problem clearer.

Thanks.
• Sep 7th 2006, 06:44 AM
CaptainBlack
Quote:

Originally Posted by tesla1410
...

BTW, is there some way I can use the trick symbols that so many others use? It would make writing a problem clearer.

The mathematicts is being laid-out/typeset using a LaTeX system, the details
can be found here.

To see what is going on left chick on an equation and a window with the
code used to set the equation will pop-up.

RonL
• Sep 7th 2006, 07:18 AM
CaptainBlack
Quote:

Originally Posted by tesla1410
Yes, you've stated the problem correctly. What's throwing me is the square root fraction w/in the square root - but then you probably know that's the hang up.

OK, simplify:

$
\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}
$

$
\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}=
\frac{1}{\sqrt{\frac{\sqrt{x+2}}{x+2}+2}}
$

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

$
=\frac{\sqrt{x+2}}{\sqrt{\sqrt{x+2}+(2x+4)}}
$

$
=\frac{\sqrt{x+2}\sqrt{\sqrt{x+2}+(2x+4)}}{\sqrt{x +2}+(2x+4)}
$

Now we need one more trick, we need:

$\left[\sqrt{A}+B\right]\left[\sqrt{A}-B\right]=A-B^2$,

so:

$
\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}=
\frac{\sqrt{x+2}\sqrt{\sqrt{x+2}+(2x+4)}}{\sqrt{x+ 2}+(2x+4)}
$
$
\times \frac{\sqrt{x+2}-(2x+4)}{\sqrt{x+2}-(2x+4)}
$

$
=\frac{\sqrt{x+2}\sqrt{\sqrt{x+2}+(2x+4)} (\sqrt{x+2}-(2x+4))}{(x+2)-(2x+4)^2}
$

I'm getting a bit tired at this point, but you can see where this is going,
and maybe I'll get back to this later or someone else may complete it if
you have further problems (also check the algebra!!).

RonL
• Sep 7th 2006, 08:01 AM
Soroban
Hello, tesla1410!

I don't blame the Cap'n for tiring out.
I suppose they want it completely rationalized . . . *sigh*

We have: . $\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}} \;= \;\frac{1}{\sqrt{\frac{\sqrt{x+2}}{x+2} + 2}}$ $= \;\frac{1}{\sqrt{\frac{\sqrt{x+2}\,+\,2(x+2)}{x+2} }}$ $= \;\sqrt{\frac{x + 2}{2(x+2)+\sqrt{x+2}}}$

Rationalize: . $\sqrt{\frac{x + 2}{2x + 4 + \sqrt{x + 2}}\cdot\frac{2x + 4 - \sqrt{x-2}}{2x+4 -\sqrt{x+2}}}$

. . . . . . . $=\;\sqrt{\frac{(x+2)(2x+4-\sqrt{x+2})}{(2x+4)^2 - (x + 2)}}$ $= \;\sqrt{\frac{(x+2)(2x+4-\sqrt{x+2})}{4x^2+15x+18}}$

Rationalize: . $\sqrt{\frac{(x+2)(2x+4 - \sqrt{x+2})}{4x^2+15x + 18}\cdot\frac{4x^2 + 15x + 18}{4x^2 + 15x + 18}}$

. . . . . . . $= \;\sqrt{\frac{(x+2)(2x+4 - \sqrt{x+2})(4x^2 + 15x + 18)}{(4x^2 + 15x + 18)^2}}$

. . . . . . . $= \;\frac{\sqrt{(x+2)(2x + 4 - \sqrt{x+2})(4x^2 + 15x + 18)}}{4x^2 + 15x + 18}$ . . . There!