1. ## Rationalising surd/simplifying denominator

Hello

I have been struggling terribly with a surd problem to the point that it's beginning to hurt }:-/

I have: √3
√2(√6-√3)
The answer is given as :
(2+√2)
2

I want to multiply the denominator by itself to clear the surds, but end up with a way too complicated answer.
I realise that

(√6-√3)(√6+√3) would rationalise part of the problem, but the √2 is really giving me a hard time...Using logic I feel I want to multiply by √2(√6+√3) but can't get anywhere near the answer if I do.

Cold someone shed some light for me please ?

Thank you!

Apologies for formatting, the question and answer are fractions of course*

2. Originally Posted by Freaked
Hello

I have been struggling terribly with a surd problem to the point that it's beginning to hurt }:-/

I have: √3
√2(√6-√3)
The answer is given as :
(2+√2)
2

I want to multiply the denominator by itself to clear the surds, but end up with a way too complicated answer.
I realise that

(√6-√3)(√6+√3) would rationalise part of the problem, but the √2 is really giving me a hard time...Using logic I feel I want to multiply by √2(√6+√3) but can't get anywhere near the answer if I do.

Cold someone shed some light for me please ?

Thank you!

Apologies for formatting, the question and answer are fractions of course*
As you say

$\displaystyle \frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})}$

$\displaystyle = \frac{\sqrt{3}}{\sqrt{2}(\sqrt{6} - \sqrt{3})} \cdot \frac{\sqrt{2}(\sqrt{6} + \sqrt{3})}{\sqrt{2}(\sqrt{6} + \sqrt{3})}$

$\displaystyle = \frac{\sqrt{3} \cdot \sqrt{2}(\sqrt{6} + \sqrt{3})}{\sqrt{2}(\sqrt{6} - \sqrt{3}) \cdot \sqrt{2}(\sqrt{6} + \sqrt{3})}$

$\displaystyle = \frac{ \sqrt{6}(\sqrt{6} + \sqrt{3})}{2(6 - 3)}$

Can you finish from here?

-Dan

3. OK

My problem then is that I couldn't simply see that √2(√6+√3) . √2(√6-√3) is straight forward...I felt that I needed to multiply all the root terms (√6-√3) by √2 as well..I will need to think about that now

Thanks : )