anonymous
  • anonymous
Can someone please help me with Algebra 2B?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ (3\sqrt{2}-2\sqrt{3})^2 }{ (3\sqrt{2})^{2}-(2\sqrt{3})^2 }\]
anonymous
  • anonymous
Rationalize the denominator and simplify.
anonymous
  • anonymous
I will try to help yikes

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tkhunny
  • tkhunny
Simplify for exploration... \(\dfrac{(a-b)^{2}}{a^{2} - b^{2}}\) Do you see that this is the same structure?
anonymous
  • anonymous
Yes I do @tkhunny
tkhunny
  • tkhunny
Can you simplify the STRUCTURE by factoring?
anonymous
  • anonymous
I don't know
tkhunny
  • tkhunny
\(\dfrac{(a-b)^{2}}{a^{2}-b^{2}} = \dfrac{(a-b)(a-b)}{(a+b)(a-b)}\) How about now?
anonymous
  • 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 ?
tkhunny
  • 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?
anonymous
  • 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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