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anonymous

  • one year ago

Help with algebra? (wait for equation in comments)

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  1. anonymous
    • one year ago
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    Rationalize the denominator and simplify. \[\frac{ 6 }{ \sqrt{2}-\sqrt{3} }\]

  2. anonymous
    • one year ago
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    You have to multiply it by 2^2+3^2/2^2+3^2, so the denominator follow this rule (a+b)(a-b)=a^2-b^2 So that the square root go away but goes upward

  3. anonymous
    • one year ago
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    Ok @gaos so now I have \[\frac{ 6(\sqrt{2}+ \sqrt{3} )}{ (\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3} }\] I know that I need to simplify the denominator to get \[(\sqrt{2})^{2}-(\sqrt{3})^2\] but I don't know what to do next

  4. anonymous
    • one year ago
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    The purpose of rationalization is to take the root out of the denominator, so now you rave to distribute 6 upward and get 2-3 downward.

  5. anonymous
    • one year ago
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    I don't understand?

  6. nikato
    • one year ago
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    when you have something like what you have in your denominator, you want to multiply by its conjugate which is just like changing the sign. ex \[\sqrt{2}-\sqrt{3} and \sqrt{2}+\sqrt{3}\]

  7. nikato
    • one year ago
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    this is so you dont have to deal with square roots in the bottom. because if you foil everything out, you get a nice number, 2-3 which equals to -1

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