# Math Help - Rearranging

1. ## Rearranging

Hey everyone, I'm new on this forum and I apologize in advance if I'm posting on the wrong sub-forum .

I'm trying to rearrange: √18/(√6+√3) to √6(√2-1).

So I got √18/[√3(√2+1)] which I cancelled down to √6/(√2+1) and then rearranged to √6(√2+1)^-1. Can anyone please tell me what the next step is? Or have I made a mistake already? Thanks.

2. Hello, and welcome !

Originally Posted by SuumEorum
Hey everyone, I'm new on this forum and I apologize in advance if I'm posting on the wrong sub-forum .
Showing your working is already a very good thing. Posting in the wrong forum would be neglectible comparing to it

I'm trying to rearrange: √18/(√6+√3) to √6(√2-1).

So I got √18/[√3(√2+1)] which I cancelled down to √6/(√2+1) |||useless ||| and then rearranged to √6(√2+1)^-1. Can anyone please tell me what the next step is? Or have I made a mistake already? Thanks.
Your working is perfect until the red message, there is the final step missing.

Remember the identity $(a-b)(a+b)=a^2-b^2$.
When you want to "simplify" a denominator containing square roots, this can be very useful because squares of square roots are (in general..) integers!

So here, try to multiply $\frac{\sqrt{6}}{\sqrt{2}+1}$ above and below (so that it's fair !) by $\sqrt{2}{\color{red}-}1$

$\frac{\sqrt{6}}{\sqrt{2}+1}=\frac{\sqrt{6}}{\sqrt{ 2}+1} \cdot \frac{\sqrt{2}{\color{red}-}1}{\sqrt{2}{\color{red}-}1} \quad \leftarrow \text{that's what I mean}$

3. Oh whoah, thanks a lot!