## anonymous one year ago Help with algebra? (wait for equation in comments)

1. anonymous

Rationalize the denominator and simplify. $\frac{ 6 }{ \sqrt{2}-\sqrt{3} }$

2. anonymous

You have to multiply it by 2^2+3^2/2^2+3^2, so the denominator follow this rule (a+b)(a-b)=a^2-b^2 So that the square root go away but goes upward

3. anonymous

Ok @gaos so now I have $\frac{ 6(\sqrt{2}+ \sqrt{3} )}{ (\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3} }$ I know that I need to simplify the denominator to get $(\sqrt{2})^{2}-(\sqrt{3})^2$ but I don't know what to do next

4. anonymous

The purpose of rationalization is to take the root out of the denominator, so now you rave to distribute 6 upward and get 2-3 downward.

5. anonymous

I don't understand?

6. nikato

when you have something like what you have in your denominator, you want to multiply by its conjugate which is just like changing the sign. ex $\sqrt{2}-\sqrt{3} and \sqrt{2}+\sqrt{3}$

7. nikato

this is so you dont have to deal with square roots in the bottom. because if you foil everything out, you get a nice number, 2-3 which equals to -1