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anonymous

  • one year ago

Can someone please help me with Algebra 2B?

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  1. anonymous
    • one year ago
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    \[\frac{ (3\sqrt{2}-2\sqrt{3})^2 }{ (3\sqrt{2})^{2}-(2\sqrt{3})^2 }\]

  2. anonymous
    • one year ago
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    Rationalize the denominator and simplify.

  3. anonymous
    • one year ago
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    I will try to help yikes

  4. tkhunny
    • one year ago
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    Simplify for exploration... \(\dfrac{(a-b)^{2}}{a^{2} - b^{2}}\) Do you see that this is the same structure?

  5. anonymous
    • one year ago
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    Yes I do @tkhunny

  6. tkhunny
    • one year ago
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    Can you simplify the STRUCTURE by factoring?

  7. anonymous
    • one year ago
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    I don't know

  8. tkhunny
    • one year ago
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    \(\dfrac{(a-b)^{2}}{a^{2}-b^{2}} = \dfrac{(a-b)(a-b)}{(a+b)(a-b)}\) How about now?

  9. anonymous
    • one year ago
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    \[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }\] so then i can cancel out the \[(3\sqrt{2}-2\sqrt{3}\] on both the top and the bottom ?

  10. tkhunny
    • one year ago
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    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
    • one year ago
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    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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