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anonymous one year ago Can someone please help me with Algebra 2B?

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1. anonymous

$\frac{ (3\sqrt{2}-2\sqrt{3})^2 }{ (3\sqrt{2})^{2}-(2\sqrt{3})^2 }$

2. anonymous

Rationalize the denominator and simplify.

3. anonymous

I will try to help yikes

4. tkhunny

Simplify for exploration... $$\dfrac{(a-b)^{2}}{a^{2} - b^{2}}$$ Do you see that this is the same structure?

5. anonymous

Yes I do @tkhunny

6. tkhunny

Can you simplify the STRUCTURE by factoring?

7. anonymous

I don't know

8. tkhunny

$$\dfrac{(a-b)^{2}}{a^{2}-b^{2}} = \dfrac{(a-b)(a-b)}{(a+b)(a-b)}$$ How about now?

9. anonymous

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

10. tkhunny

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?

11. anonymous

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.

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