Question

Help with algebra? (wait for equation in comments)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

I don't understand?

Answer 6

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 \[\sqrt{2}-\sqrt{3} and \sqrt{2}+\sqrt{3}\]

Answer 7

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