The problem is just to find the distance between ( $- \sqrt{2} , - \sqrt{6}$) and ( $\sqrt{3} , - \sqrt{2}$)

I get;
$\sqrt{13+ 2\sqrt{2}\sqrt{3}-2\sqrt{6}\sqrt{2}}$

while the book shows;

$\sqrt{13+2\sqrt{6}-4\sqrt{3}}$

I simplified the problems on a ti-30 and got the same decimal number for each answer.
So I think I'm just looking for a way to convert between the two forms.

2. When you are multiplying roots, you merely multiply what is under the root sign. sqrt(2)*sqrt(3) = sqrt(6). Additionally, sqrt(6) can be simplified to 2*sqrt(3). Can you take it from there?

3. Originally Posted by profound
Additionally, sqrt(6) can be simplified to 2*sqrt(3).
You mean sqrt(12), not sqrt(6), right?

4. Ah OK. So the $-2\sqrt{6}\sqrt{2} = -2\sqrt{12} = -2\sqrt{4}\sqrt{3} = -4\sqrt{3}$

Is this because of the commutative property?

(thanks btw)

5. Originally Posted by bkbowser
Ah OK. So the $-2\sqrt{6}\sqrt{2} = -2\sqrt{12} = -2\sqrt{4}\sqrt{3} = \boxed{-2 \cdot 2 \sqrt{3}} =-4\sqrt{3}$

Is this because of the commutative property?

(thanks btw)
Hi bkbowser,

It's not the commutative property that you're using here.

The commutative property has to do with the order in addition and multiplication.

Addition: a + b = b + a

Multiplication: ab = ba

You simply extracted the square root of the largest perfect square factor of 12 and multiplied it by the factor -2 to arrive at your -4 at the end there.

6. This probably seems silly but, presumably I was taught how too do this specific operation before. And then promptly forgot that it exists. So I'd just like to over learn as much about it as I can so that I remember it for later. So knowing anything extraneous would be helpful.

So far all I can think of is something like; $2\sqrt{6}\sqrt{2}$ implies $2\sqrt{a+b}\sqrt{c-d} = 2\sqrt{ac-ad+bc-bd}$; as some thing additional to hold onto.

7. It is more the "associative property"- a(bc)= (ab)c.