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: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
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.
OK, simplify: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
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!