1. ## square root problem

hello guys, i got a problem in solving this

if
$x=\tfrac{2ab}{{b_{}}^{2}+1}
$

calculate :
$\tfrac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}$

thanks

2. Uhm, just multiply by $\sqrt{a+x}-\sqrt{a-x}$ and then calculate. Where is the problem? o.o

3. i know that, i got stucked on this :

$\frac{2a-2\sqrt{a^{2}-x^{2}}}{2x}$

4. And then calculate, remember? Replace x by $2ab/(b^2+1)$.

5. Originally Posted by disharmonie
hello guys, i got a problem in solving this

if
$x=\tfrac{2ab}{{b_{}}^{2}+1}
$

calculate :
$\tfrac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}$

thanks
Why not just calculate the 2 relevant terms first to see what you are dealing with ?

$a+x=a+\frac{2ab}{b^2+1}=\frac{a\left(b^2+1\right)} {b^2+1}+\frac{2ab}{b^2+1}$

$=\frac{ab^2+a+2ab}{b^2+1}=\frac{a\left(b^2+2b+1\ri ght)}{b^2+1}$

$=\frac{a(b+1)^2}{b^2+1}$

Then

$\sqrt{a+x}=(b+1)\sqrt{\frac{a}{b^2+1}}$

Similarly

$\sqrt{a-x}=(b-1)\sqrt{\frac{a}{b^2+1}$

which gives

$\sqrt{a+x}-\sqrt{a-x}=2\sqrt{\frac{a}{b^2+1}}$

$\sqrt{a+x}+\sqrt{a-x}=2b\sqrt{\frac{a}{b^2+1}}$

Then it's quite straightforward to perform the division.

6. just replacing x with will lead into a complex calculation. thanks to Archie, that's a better method to solve this