I've been asked to simplify and I don't know how.

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

Wow.

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- Sep 6th 2006, 08:10 PMtesla1410Simplifying 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 PMCaptainBlackQuote:

Originally Posted by**tesla1410**

$\displaystyle

\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:

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

RonL - Sep 7th 2006, 06:35 AMtesla1410
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 AMCaptainBlackQuote:

Originally Posted by**tesla1410**

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 AMCaptainBlackQuote:

Originally Posted by**tesla1410**

$\displaystyle

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

$

Start with the innermost square root:

$\displaystyle

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

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

$

$\displaystyle

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

$

$\displaystyle

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

$

$\displaystyle

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

$

Now we need one more trick, we need:

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

so:

$\displaystyle

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

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

$$\displaystyle

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

$

$\displaystyle

=\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 AMSoroban
Hello, tesla1410!

I don't blame the Cap'n for tiring out.

I suppose they want it__completely__rationalized . . . *sigh*

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

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

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

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

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

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