Question

Can someone please help me with Algebra 2B?

Answer 1

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

Answer 2

Rationalize the denominator and simplify.

Answer 3

I will try to help yikes

Answer 4

Simplify for exploration...

\(\dfrac{(a-b)^{2}}{a^{2} - b^{2}}\)

Do you see that this is the same structure?

Answer 5

Yes I do @tkhunny

Answer 6

Can you simplify the STRUCTURE by factoring?

Answer 7

I don't know

Answer 8

\(\dfrac{(a-b)^{2}}{a^{2}-b^{2}} = \dfrac{(a-b)(a-b)}{(a+b)(a-b)}\)

How about now?

Answer 9

\[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }\] so then i can cancel out the \[(3\sqrt{2}-2\sqrt{3}\] on both the top and the bottom ?

Answer 10

Okay, that was just a hint. This one turns out to be easier because "rationalize the denominator" doesn't actually mean anything for this problem. It's ALREADY rational!!

\(\left(3\sqrt{2}\right)^{2} - \left(2\sqrt{3}\right)^{2} = 9\cdot 2 - 4\cdot 3 = 18 - 12 = 6\)

How's that for Rational?

Answer 11

OK but @tkhunny the original problem wasn't actually rational the problem I gave was just the step I was at. I need to get it so there is no denominator. The original problem was \[\frac{ 3\sqrt{2}-2\sqrt{3} }{ 3\sqrt{2}+2\sqrt{3} }\] and then I multiplied by the conjugate. So if what we have right now is \[\frac{ (3\sqrt{2}-2\sqrt{3})^2 }{ 6 }\] (if I understood you correctly) then I also need to get rid of the six but I don't know how.