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Math Help - Multiply Two Factors

  1. #1
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    Multiply Two Factors

    I am just looking through my course notes right now.

    I notice the prof went to multiply two terms and I am kind of lost on the method he used here.

    A problem here is that I don't know how to show root of with my keyboard LOL. I am sure someone will

    (a+b√2)(c+d√2)=(ac+2bd)+(ad+bc)√2

    Is this correct or did I copy something down wrong?
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  2. #2
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    re: Multiply Two Factors

    Quote Originally Posted by ehpoc View Post
    I am just looking through my course notes right now.

    I notice the prof went to multiply two terms and I am kind of lost on the method he used here.

    A problem here is that I don't know how to show root of with my keyboard LOL. I am sure someone will

    (a+b√2)(c+d√2)=(ac+2bd)+(ad+bc)√2

    Is this correct or did I copy something down wrong?
    Do you know how to expand brackets using FOIL?
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  3. #3
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    re: Multiply Two Factors

    [tex]\sqrt{2}[/tex] produces \sqrt{2}.

    Quote Originally Posted by ehpoc View Post
    (a+b√2)(c+d√2)=(ac+2bd)+(ad+bc)√2

    Is this correct or did I copy something down wrong?
    This is correct. Just use distributivity and note that \sqrt{2}\sqrt{2}=2.
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  4. #4
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    re: Multiply Two Factors

    Ok there were two reasons why this confused me.

    1. the 2bd which should be clear to me.....ooops

    2.In my notes where there should have been a "c" where there was an "a" so I was getting a^2 haha ...math in the morning.

    But now I have another similar related question about some simple algebra

    I have

    (a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?
    EDIT: For this one I should have added context. Because from a similar proof it removes the √2 because a and b are rational. I still dont get the nest part though...

    From that he went to a^2-2b^2 Not sure what happened here.
    Last edited by ehpoc; December 14th 2011 at 07:35 AM.
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  5. #5
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    re: Multiply Two Factors

    Quote Originally Posted by ehpoc View Post
    (a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?
    I am not sure about this equation. I can only say that

    \frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a-b\sqrt{2}}{(a-b\sqrt{2})(a+b\sqrt{2})}=\frac{1}{a+b\sqrt{2}}
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  6. #6
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    Re: Multiply Two Factors

    You can see it in contest in this link.

    Look under the heading "Inverses"

    Numbers of Type Rational a plus b root 2 Form a Field - ProofWiki
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  7. #7
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    Re: Multiply Two Factors

    Quote Originally Posted by ehpoc View Post
    (a-b√2)/(a^2-2b^2) =( a/a-2b^2)-(b/a^2+2b^2) where did the √2 go?
    I don't see this equation in the link you provided. The link says that the inverse of a+b\sqrt{2}, which is \left({\dfrac {a - b \sqrt 2} {a^2 - 2b^2}}\right), equals \dfrac a {a^2 - 2b^2} - \dfrac {b \sqrt 2} {a^2 - 2b^2}, with \sqrt{2} in the nominator of the second term. The coefficients \dfrac a {a^2 - 2b^2} and \dfrac {b} {a^2 - 2b^2} are rational number when a and b are.

    By the way, according to the standard arithmetic operation precedence, a/a-2b^2 means (a/a)-2b^2 = 1 - 2b^2. It should probably be a/(a^2-2b^2).
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  8. #8
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    Re: Multiply Two Factors

    The equation in my notes is doing the same proof same proof as the link I provided. In that one too the √2 seems to disappear. Whatever though I will just go see my prof about this.
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