Question

Help with Algebra 2B? Rationalize the denominator and simplify (wait for the equation in comments):

Answer 1

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

Answer 2

@Hero

Answer 3

Multiply top and bottom by the conjugate.

Answer 4

So @Hero it would be \[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }\]
which would simplify to \[\frac{ 3\sqrt{2}^{2}+2\sqrt{3}^{2} }{ 3\sqrt{2}^{2}-2\sqrt{3}^{2} }\] am I right so far?

Answer 5

It's more like:

\[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) } = \frac{(3\sqrt{2}-2\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2}\]

Answer 6

Oh ok thanks that helps a lot I think I can do it from here.